bzoj2252 矩阵距离

很好奇这种 普及- 的题为什么会是权限题......

我一开始想用枚举 + 搜索，看书后发现自己脑抽了。直接BFS即可。

 #include <cstdio>
 #include <queue>
 #include <cstring>

 struct Sta {
     int x, y, step;
     Sta(int x = , int y = ) {
         this->x = x;
         this->y = y;
     }
 };

 const int N = ;
 int G[N][N], dis[N][N];

 const int dx[] = {, , , -};
 const int dy[] = {, -, , };

 int main() {
     int m, n;
     std::queue<Sta> Q;
     memset(dis, -, sizeof(dis));
     scanf("%d%d", &n, &m);
     for(int i = ; i <= n; i++) {
         for(int j = ; j <= m; j++) {
             char c = getchar();
             while(c < '' || c > '') {
                 c = getchar();
             }
             G[i][j] = c - ;
             if(G[i][j]) {
                 dis[i][j] = ;
                 Q.push(Sta(i, j));
             }
         }
     }

     for(int i = ; i <= n + ; i++) {
         dis[i][] = dis[i][m + ] = ;
     }
     for(int i = ; i <= m; i++) {
         dis[][i] = dis[n + ][i] = ;
     }

     while(!Q.empty()) {
         Sta op = Q.front();
         Q.pop();
         for(int i = ; i <= ; i++) {
             int tx = op.x + dx[i];
             int ty = op.y + dy[i];
             if(dis[tx][ty] >= ) {
                 continue;
             }
             dis[tx][ty] = dis[op.x][op.y] + ;
             Q.push(Sta(tx, ty));
         }
     }

     for(int i = ; i <= n; i++) {
         for(int j = ; j <= m; j++) {
             printf("%d ", dis[i][j]);
         }
         if(i < n) {
             puts("");
         }
     }

     return ;
 }

AC代码

代码里的G[N][N]应该是多余的...