kuangbin专题一简单搜索-白红宇

kuangbin专题一简单搜索

阅读量：7125 次

发布时间：2019-06-28

本文共 22324 字，大约阅读时间需要 74 分钟。

弱菜做了好久23333333........

传送门：

A 只能摆k个棋子，只能摆在#

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 using namespace std; 5 const int max_n = 10; 6 char maze[max_n][max_n]; 7 bool vis[max_n]; 8 int n, m, ans, cnt; 9 void init(){10     memset(maze, 0, sizeof(maze));11     ans = cnt = 0;12 }13 void dfs(int k){14     if(cnt == m){ ++ans; return ; }15     if(k > n) return ;16     for(int i = 0; i < n; ++i){17         if(maze[i][k] == '#' && !vis[i]){18             vis[i] = 1; ++cnt;19             dfs(k+1);20             vis[i] = 0; --cnt;21         }22     }23     dfs(k+1);24 }25 int main(){26     ios::sync_with_stdio(false);27     while(cin >> n >> m){28         if(n == m && n == -1) break;29         init();30         for(int i = 0; i < n; ++i){31             cin.get();32             for(int j = 0; j < n; ++j) cin >> maze[i][j];33         }34         dfs(0);35         cout << ans << endl;36     }37     return 0;38 }

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B 三维BFS

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 #include 
         
           5 #include 
          
            6 using namespace std; 7 const int max_n = 40; 8 const int cx[] = {
    0, 1, 0, -1, 0, 0}; 9 const int cy[] = {
    1, 0, -1, 0, 0, 0};10 const int cz[] = {
    0, 0, 0, 0, 1, -1};11 int maze[max_n][max_n][max_n];12 bool vis[max_n][max_n][max_n];13 struct point{14     int x, y, z, step;15 };16 int L, R, C;17 int ex, ey, ez;18 int main(){19     ios::sync_with_stdio(false);20     while(cin >> L >> R >> C){21         if(L == 0 && R == 0 && C == 0) break;22         queue
           
             q;23 memset(vis, 0, sizeof(vis));24 memset(maze, 0, sizeof(maze));25 for(int k = 0; k < L; ++k){26 for(int i = 0; i < R; ++i){27 cin.get();28 for(int j = 0; j < C; ++j){29 char ch = cin.get();30 if(ch == 'S'){31 maze[k][i][j] = 1;32 point p = {k, i, j, 0};33 q.push(p);34 }35 else if(ch == '.') maze[k][i][j] = 1;36 else if(ch == 'E'){37 maze[k][i][j] = 1;38 ex = k; ey = i; ez = j;39 }40 }41 }42 cin.get();43 }44 int x, y, z, step;45 while(!q.empty()){46 point p = q.front();47 q.pop();48 x = p.x, y = p.y, z = p.z, step = p.step;49 if(x == ex && y == ey && z == ez) break;50 for(int i = 0; i < 6; ++i){51 int nx = x + cx[i], ny = y + cy[i], nz = z + cz[i];52 if(nx<0||L
            
             <0||R
             
              <0||C

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C 搜......

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 #include 
         
           5 #include 
          
            6 using namespace std; 7 struct point{ 8     int x, step; 9 };10 int S, E;11 int vis[100010];12 int main(){13     ios::sync_with_stdio(false);14     while(cin >> S >> E){15         memset(vis, 0, sizeof(vis));16         vis[S] = 1;17         queue
           
             q;18 q.push(point{S, 0});19 int x, step;20 while(!q.empty()){21 point p = q.front();22 x = p.x; step = p.step;23 q.pop();24 if(x == E) break;25 for(int i = 0; i < 3; ++i){26 int nx;27 if(i == 0) nx = x + 1;28 else if(i == 1) nx = x - 1;29 else nx = x << 1;30 if(0 <= nx && nx <= 100000 && !vis[nx]){ q.push(point{nx, step+1}); vis[nx] = 1; }31 }32 }33 cout << step << endl;34 }35 return 0;36 }

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D 卡了好久看了题解第一行确定下来后其他行的也都确定下来了(说的好有道理)，所以枚举第一行状态，确定下其他行，看最后一行是否都是0

　　然后在这道题发现了之前二进制枚举的一个错误....... 正确姿势：(i>>j) & 1（吧？）

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 #include 
         
           5 #include 
          
            6 using namespace std; 7  8 const int MAXN = 20; 9 const int INF = 0x3f3f3f3f;10 const int cx[] = {
    0, 0, -1, 0};11 const int cy[] = {
    1, -1, 0, 0};12 int n, m;13 int ans[MAXN][MAXN];14 int flip[MAXN][MAXN];15 int origin[MAXN][MAXN];16 17 int check(int x, int y){18     int cnt = origin[x][y];19     for(int i = 0; i < 4; ++i){20         int nx = x + cx[i], ny = y + cy[i];21         if(nx >= 0 && nx < n && ny >= 0 && ny < m)22             cnt += flip[nx][ny];23     }24     return cnt & 1;25 }26 27 int duang(){28     for(int i = 1; i < n; ++i)29         for(int j = 0; j < m; ++j)30             if(check(i-1, j)) flip[i][j] = 1;31     for(int j = 0; j < m; ++j)32         if(check(n-1, j)) return 0;33     int cnt = 0;34     for(int i = 0; i < n; ++i)35         for(int j = 0; j < m; ++j)36             cnt += flip[i][j];37     return cnt;38 }39 40 int main(){41     while(cin >> n >> m){42         memset(origin, 0, sizeof(origin));43         for(int i = 0; i < n; ++i)44             for(int j = 0; j < m; ++j)45                 cin >> origin[i][j];46         int tans = INF;47         for(int i = 0; i < (1<
           
            >j) & 1;51 int cnt = duang();52 if(cnt < tans && cnt != 0){53 tans = cnt;54 memcpy(ans, flip, sizeof(flip));55 }56 }57 if(tans == INF) cout << "IMPOSSIBLE" << endl;58 else{59 for(int i = 0; i < n; ++i)60 for(int j = 0; j < m; ++j)61 cout << ans[i][j] << " \n"[j==m-1];62 }63 }64 return 0;65 }

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E 听说答案都在long long范围内hhhh.......

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 using namespace std; 5 typedef long long ll; 6 int main(){ 7     ios::sync_with_stdio(false); 8     int n; 9     while(cin >> n){10         if(n == 0) break;11         queue
         
           q;12         q.push(1LL);13         while(!q.empty()){14             ll k = q.front();15             q.pop();16             if(k % n == 0){17                 cout << k << endl;18                 break;19             }20             q.push(k*10);21             q.push(k*10+1);22         }23     }24     return 0;25 }

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F GOJ有这道题........

1 #include  2 #include 
       
         3 #include 
        
          4 #include 
         
           5 #include 
          
            6 #include 
           
             7 #include 
            
              8 #include 
             
               9 #include 
              
               10 #include 
               
                11 #include 
                
                 12 #include 
                 
                  13 #include 
                  
                   14 using namespace std;15 int m, n;16 bool isPrime[10010];17 int vis[10010];18 void preprocess(){19 memset(isPrime, true, sizeof(isPrime));20 isPrime[0] = isPrime[1] = false;21 for(int i = 2; i < 10000; ++i){22 if(isPrime[i]){23 for(int j = i; i*j < 10000; ++j){24 isPrime[i*j] = false;25 }26 }27 }28 }29 void bfs(){30 memset(vis, 0, sizeof(vis));31 vis[m] = 1;32 queue
                   
                     q;33 q.push(m);34 while(!q.empty()){35 int k = q.front();36 q.pop();37 int d[4] = {k/1000, k%1000/100, k%100/10, k%10};38 int b[4] = { 1000, 100, 10, 1};39 for(int i = 0; i < 4; ++i){40 for(int j = 0; j < 10; ++j){41 int t = k - d[i] * b[i] + j * b[i];42 if(k == n) return ;43 if(t > 999 && isPrime[t] && !vis[t]){44 vis[t] = vis[k] + 1;45 q.push(t);46 }47 }48 }49 }50 return ;51 }52 int main(){53 ios::sync_with_stdio(false);54 preprocess();55 int t;56 cin >> t;57 while(t--){58 cin >> m >> n;59 bfs();60 if(vis[n]) cout << vis[n]-1 << endl;61 else cout << "Impossible" << endl;62 }63 return 0;64 }

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G 水直接模拟略坑是第一张都是最底的一张

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 #include 
         
           5 using namespace std; 6  7  8 int main(){ 9     int t;10     cin >> t;11     for(int x = 1; x <= t; ++x){12         int l;13         cin >> l;14         string str1, str2, desstr;15         cin >> str1 >> str2 >> desstr;16         int step = 0;17         set
          
            s;18         s.insert(str1+str2);19         bool flag = true;20         while(flag){21             string tmpstr;22             for(int i = 0; i < l; ++i){23                 tmpstr += str2[i];24                 tmpstr += str1[i];25             }26             //cout << tmpstr << endl;27             ++step;28             if(tmpstr == desstr) break;29             if(s.count(tmpstr)){30                 flag = false;31                 break;32             }33             s.insert(tmpstr);34             str1 = tmpstr.substr(0, l);35             str2 = tmpstr.substr(l, 2*l);36             //cout << str1 << " "  << str2 << endl;37         }38         cout << x << ' ' << (flag ? step : -1) << endl;39     }40     return 0;41 }

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H M题写吐了H不想做.......

还是搜 6种状态

I 暴力找两个#搜.......

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 #include 
         
           5 #include 
          
            6 using namespace std; 7  8 const int INF =  0x3f3f3f; 9 const int cx[4] = {
    0, 0, 1, -1};10 const int cy[4] = {
    1, -1, 0, 0};11 int n, m;12 int dis[15][15];13 char map[15][15];14 15 struct Point{16     int x, y;17     Point(){}18     Point(int xx, int yy) :x(xx), y(yy){}19 };20 queue
           
             q;21 int bfs(int x1, int y1, int x2, int y2){22 memset(dis, INF, sizeof(dis));23 q.push(Point(x1, y1));24 q.push(Point(x2, y2));25 dis[x1][y1] = 0;26 dis[x2][y2] = 0;27 while (!q.empty()){28 Point p = q.front();29 q.pop();30 for (int i = 0; i < 4; i++){31 int xx = p.x + cx[i], yy = p.y + cy[i];32 if (xx >= 0 && xx < n && yy >= 0 && yy
            
              dis[p.x][p.y] + 1){33 dis[xx][yy] = dis[p.x][p.y] + 1;34 q.push(Point(xx, yy));35 }36 }37 }38 39 int ret = 0;40 for (int i = 0; i < n; i++)41 for (int j = 0; j < m; j++)42 if (map[i][j] == '#')43 ret = max(ret, dis[i][j]);44 return ret;45 }46 47 int main(){48 int t;49 cin >> t;50 for (int x = 1; x <= t; x++){51 while(!q.empty()) q.pop();52 cin >> n >> m;53 for (int i = 0; i < n; i++)54 cin >> map[i];55 int ans = INF;56 for (int i = 0; i < n; i++)57 for (int j = 0; j < m; j++)58 if (map[i][j] == '#')59 for (int ii = 0; ii < n; ii++)60 for (int jj = 0; jj < m; jj++)61 if (map[ii][jj] == '#'){62 int temp = bfs(i, j, ii, jj);63 ans = min(ans, temp);64 }65 if (ans == INF) ans = -1;66 cout << "Case " << x << ": " << ans << endl;67 }68 return 0;69 }

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J 多处火.....先处理火，再处理人，遇到可行的情况退出来

1 #include 
      
        2 using namespace std; 3  4 struct pos{ 5     int x, y; 6     pos(int xx = 0, int yy = 0): x(xx), y(yy) {} 7 }; 8 const int MAXN = 1111; 9 const int INF = 0x7f7f7f7f;10 const int cx[] = {
    0, 0, 1, -1};11 const int cy[] = {
    1, -1, 0, 0};12 int n, m;13 int jx, jy, fx, fy;14 int maze[MAXN][MAXN];15 int jstep[MAXN][MAXN];16 int fstep[MAXN][MAXN];17 18 int main(){19     int t;20     cin >> t;21     while(t--){22         cin >> n >> m;23         queue
       
         qf;24         memset(fstep, -1, sizeof(fstep));25         for(int i = 0; i < n; ++i){26             cin.get();27             for(int j = 0; j < m; ++j){28                 char ch;29                 cin >> ch;30                 if(ch == '#') maze[i][j] = 0;31                 else{32                     maze[i][j] = 1;33                     if(ch == 'J'){34                         jx = i; 35                         jy = j;36                     }37                     else if(ch == 'F'){38                         fx = i;39                         fy = j;40                         qf.push(pos(fx, fy));41                         fstep[fx][fy] = 0;42                     }43                 }44             }45         }46         while(!qf.empty()){47             pos p = qf.front();48             int x = p.x, y = p.y;49             qf.pop();50             for(int i = 0; i < 4; ++i){51                 int nx = x + cx[i], ny = y + cy[i];52                 if(nx < 0 || n <= nx || ny < 0 || m <= ny || maze[nx][ny] == 0 || fstep[nx][ny] != -1) continue;53                 qf.push(pos(nx, ny));54                 fstep[nx][ny] = fstep[x][y] + 1;55             }56         }57         // cout << endl;58         // for(int i = 0; i < n; ++i){59         //     for(int j = 0; j < m; ++j)60         //         cout << fstep[i][j] << '\t';61         //     cout << endl;62         // }63         memset(jstep, -1, sizeof(jstep));64         queue
        
          qj;65         qj.push(pos(jx, jy));66         jstep[jx][jy] = 0;67         int ans = -1;68         while(!qj.empty()){69             pos p = qj.front();70             int x = p.x, y = p.y;71             qj.pop();72             if(x == 0 || x == n-1 || y == 0 || y == m-1){73                 ans = jstep[x][y];74                 break;75             }76             for(int i = 0; i < 4; ++i){77                 int nx = x + cx[i], ny = y + cy[i];78                 if(nx < 0 || n <= nx || ny < 0 || m <= ny || maze[nx][ny] == 0 || jstep[nx][ny] != -1 || (fstep[nx][ny] != -1 && fstep[nx][ny] <= jstep[x][y] + 1)) continue;79                 qj.push(pos(nx, ny));80                 jstep[nx][ny] = jstep[x][y] + 1;81             }82         }83         // cout << endl;84         // for(int i = 0; i < n; ++i){85         //     for(int j = 0; j < m; ++j)86         //         cout << jstep[i][j] << '\t';87         //     cout << endl;88         // }89         if(ans == -1) cout << "IMPOSSIBLE" << endl;90         else cout << ans+1 << endl;91     }92     return 0;93 }

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K 搜开数组记录上一个点的坐标

1 #include  2 #include 
       
         3 #include 
        
          4 #include 
         
           5 #include 
          
            6 #include 
           
             7 #include 
            
              8 #include 
             
               9 #include 
              
               10 #include 
               
                11 #include 
                
                 12 #include 
                 
                  13 #include 
                  
                   14 #include 
                   
                    15 using namespace std;16 struct point{17 int x, y;18 };19 int maze[5][5];20 point state[5][5];21 const int cx[] = { 0, 0, 1, -1};22 const int cy[] = { 1, -1, 0, 0};23 int main(){24 ios::sync_with_stdio(false);25 memset(state, 0, sizeof(state));26 for(int i = 0; i < 5; ++i)27 for(int j = 0; j < 5; ++j)28 cin >> maze[i][j];29 queue
                    
                      q;30 q.push({ 0, 0});31 maze[0][0] = 1;32 bool flag = false;33 while(!q.empty()){34 point k = q.front();35 q.pop();36 for(int i = 0; i < 4; ++i){37 int x = k.x + cx[i], y = k.y + cy[i];38 if(x < 0 || x > 4 || y < 0 || y > 4 || maze[x][y]) continue;39 q.push({x, y});40 maze[x][y] = 1;41 state[x][y] = k;42 if(x == 4 && y == 4){43 flag = true;44 break;45 }46 }47 if(flag) break;48 }49 point k = { 4, 4};50 stack
                     
                       s;51 while(!(k.x == 0 && k.y == 0)){52 s.push(k);53 k = state[k.x][k.y];54 }55 s.push(point{ 0, 0});56 while(!s.empty()){57 cout << "(" << s.top().x << ", " << s.top().y << ")" << endl;58 s.pop();59 }60 return 0;61 }

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L 数水坑类的 8个方向

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 using namespace std; 5 int ans; 6 char maze[120][120]; 7 const int cx[] = {
    0, 0, 1, 1, 1, -1, -1, -1}; 8 const int cy[] = {
    1, -1, 1, -1, 0, 1, 0, -1}; 9 void dfs(int x, int y){10     maze[x][y] = '*';11     for(int i = 0; i < 8; ++i){12         int nx = x + cx[i], ny = y + cy[i];13         if(maze[nx][ny] == '@') dfs(nx, ny);14     }15 }16 int main(){17     ios::sync_with_stdio(false);18     int m, n;19     while(cin >> m >> n){20         if(m == 0 && n == 0) break;21         ans = 0;22         for(int i = 0; i < m; ++i){23             cin.get();24             for(int j = 0; j < n; ++j)25                 cin >> maze[i][j];26         }27         for(int i = 0; i < m; ++i){28             for(int j = 0; j < n; ++j){29                 if(maze[i][j] == '@'){30                     ++ans;31                     dfs(i, j);32                 }33             }34         }35         cout << ans << endl;36     }37     return 0;38 }

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M 可乐.....改了好久蜜汁哇果断重写简直恶心

1 #include 
      
         2 using namespace std;  3 //#define print() cout<
       
        <<" "<
        
         <<" "<
         
          <
          
            q; 11 const int MAXN = 111; 12 int step[MAXN][MAXN][MAXN]; 13  14 int main(){ 15     while(cin >> s >> n >> m){ 16         if(s == 0 && n == 0 && m == 0) break; 17         if(s & 1) cout << "NO" << endl; 18         else{ 19             while(!q.empty()) q.pop(); 20             memset(step, -1, sizeof(step)); 21             int ans = -1; 22             q.push(state(s, 0, 0)); 23             step[s][0][0] = 0; 24             while(!q.empty()){ 25                 state p = q.front(); 26                 q.pop(); 27                 int ss = p.s, nn = p.n, mm = p.m; 28                 if((ss == s/2 && nn == s/2) || (ss == s/2 && mm == s/2) || (nn == s/2 && mm == s/2)){ 29                     ans = step[ss][nn][mm]; 30                     break; 31                 } 32                 state tmp; 33                 //s -> n || s -> m 34                 if(ss){ 35                     //s -> n 36                     if(ss > n - nn){ 37                         tmp.s = ss - (n - nn); 38                         tmp.n = n; 39                         tmp.m = mm; 40                     } 41                     else{ 42                         tmp.s = 0; 43                         tmp.n = nn + ss; 44                         tmp.m = mm; 45                     } 46                     if(step[tmp.s][tmp.n][tmp.m] == -1){ 47                         step[tmp.s][tmp.n][tmp.m] = step[ss][nn][mm] + 1; 48                         q.push(tmp); 49                     } 50                     //ss -> m 51                     if(ss > m - mm){ 52                         tmp.s = ss - (m - mm); 53                         tmp.m = m; 54                         tmp.n = nn; 55                     } 56                     else{ 57                         tmp.s = 0; 58                         tmp.m = mm + ss; 59                         tmp.n = nn; 60                     }     61                     if(step[tmp.s][tmp.n][tmp.m] == -1){ 62                         step[tmp.s][tmp.n][tmp.m] = step[ss][nn][mm] + 1; 63                         q.push(tmp); 64                     } 65                 } 66                 //n -> s || n -> m 67                 if(nn){ 68                     //n -> s 69                     if(nn > s - ss){ 70                         tmp.n  = nn - (s - ss); 71                         tmp.s = s; 72                         tmp.m = mm; 73                     } 74                     else{ 75                         tmp.n = 0; 76                         tmp.s = ss + nn; 77                         tmp.m = mm; 78                     } 79                     if(step[tmp.s][tmp.n][tmp.m] == -1){ 80                         step[tmp.s][tmp.n][tmp.m] = step[ss][nn][mm] + 1; 81                         q.push(tmp); 82                     } 83                     //n -> m 84                     if(nn > m - mm){ 85                         tmp.n  = nn - (m - mm); 86                         tmp.m = m; 87                         tmp.s = ss; 88                     } 89                     else{ 90                         tmp.n = 0; 91                         tmp.m = mm + nn; 92                         tmp.s = ss; 93                     } 94                     if(step[tmp.s][tmp.n][tmp.m] == -1){ 95                         step[tmp.s][tmp.n][tmp.m] = step[ss][nn][mm] + 1; 96                         q.push(tmp); 97                     } 98                 } 99                 //m -> n || m -> s100                 if(mm){101                     //m -> n102                     if(mm > n - nn){103                         tmp.m = mm - (n - nn);104                         tmp.n = n;105                         tmp.s = ss;106                     }107                     else{108                         tmp.m = 0;109                         tmp.n = nn + mm;110                         tmp.s = ss;111                     }112                     if(step[tmp.s][tmp.n][tmp.m] == -1){113                         step[tmp.s][tmp.n][tmp.m] = step[ss][nn][mm] + 1;114                         q.push(tmp);115                     }116                     //m -> s117                     if(mm > s - ss){118                         tmp.m = mm - (s - ss);119                         tmp.s = s;120                         tmp.n = nn;121                     }122                     else{123                         tmp.m = 0;124                         tmp.s = mm + ss;125                         tmp.n = nn;126                     }127                     if(step[tmp.s][tmp.n][tmp.m] == -1){128                         step[tmp.s][tmp.n][tmp.m] = step[ss][nn][mm] + 1;129                         q.push(tmp);130                     }131                 }132             }133             if(ans == -1) cout << "NO" << endl;134             else cout << ans << endl;135         }136     }137     return 0;138 }

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N 从两个人的位置开始bfs整个图

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 #include 
         
           5 #include 
          
            6 #include 
           
             7 #include 
            
              8 using namespace std; 9 10 struct pos{11 int x, y;12 pos(int xx, int yy): x(xx), y(yy){};13 };14 const int MAXN = 222;15 char maze[MAXN][MAXN];16 const int cx[] = { 1, -1, 0, 0};17 const int cy[] = { 0, 0, 1, -1};18 vector
             
               kfc;19 queue
              
                qy; int yx, yy, stepy[MAXN][MAXN];20 queue
               
                 qm; int mx, my, stepm[MAXN][MAXN];21 22 int main(){23 int n, m;24 while(cin >> n >> m){25 kfc.clear();26 for(int i = 0; i < n; ++i){27 cin.get();28 for(int j = 0; j < m; ++j){29 char ch;30 cin >> ch;31 if(ch == '#') maze[i][j] = ch;32 else{33 if(ch == 'Y'){34 yx = i;35 yy = j;36 }37 if(ch == 'M'){38 mx = i;39 my = j;40 }41 if(ch == '@')42 kfc.push_back(pos(i, j));43 maze[i][j] = '.';44 }45 }46 }47 while(!qy.empty()) qy.pop();48 while(!qm.empty()) qm.pop();49 memset(stepy, -1, sizeof(stepy));50 memset(stepm, -1, sizeof(stepm));51 qy.push(pos(yx, yy));52 stepy[yx][yy] = 0;53 while(!qy.empty()){54 int x = qy.front().x, y = qy.front().y;55 qy.pop();56 for(int i = 0; i < 4; ++i){57 int nx = x + cx[i], ny = y + cy[i];58 if(nx < 0 || n < nx || ny < 0 || m < ny || maze[nx][ny] == '#' || stepy[nx][ny] != -1) continue;59 stepy[nx][ny] = stepy[x][y] + 1;60 qy.push(pos(nx, ny));61 }62 }63 qm.push(pos(mx, my));64 stepm[mx][my] = 0;65 while(!qm.empty()){66 int x = qm.front().x, y = qm.front().y;67 qm.pop();68 for(int i = 0; i < 4; ++i){69 int nx = x + cx[i], ny = y + cy[i];70 if(nx < 0 || n < nx || ny < 0 || m < ny || maze[nx][ny] == '#' || stepm[nx][ny] != -1) continue;71 stepm[nx][ny] = stepm[x][y] + 1;72 qm.push(pos(nx, ny));73 }74 }75 int ans = 0x7f7f7f7f;76 for(vector
                
                 ::iterator i = kfc.begin(); i != kfc.end(); ++i){77 int x = (*i).x, y = (*i).y;78 ans = min(ans, stepy[x][y] + stepm[x][y]);79 }80 cout << ans*11 << endl;81 }82 return 0;83 }