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 = 0, int y = 0) {
        this->x = x;
        this->y = y;
    }
};

const int N = 1010;
int G[N][N], dis[N][N];

const int dx[4] = {0, 0, 1, -1};
const int dy[4] = {1, -1, 0, 0};

int main() {
    int m, n;
    std::queue<Sta> Q;
    memset(dis, -1, sizeof(dis));
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= m; j++) {
            char c = getchar();
            while(c < '0' || c > '1') {
                c = getchar();
            }
            G[i][j] = c - 48;
            if(G[i][j]) {
                dis[i][j] = 0;
                Q.push(Sta(i, j));
            }
        }
    }

    for(int i = 0; i <= n + 1; i++) {
        dis[i][0] = dis[i][m + 1] = 0;
    }
    for(int i = 1; i <= m; i++) {
        dis[0][i] = dis[n + 1][i] = 0;
    }

    while(!Q.empty()) {
        Sta op = Q.front();
        Q.pop();
        for(int i = 0; i <= 3; i++) {
            int tx = op.x + dx[i];
            int ty = op.y + dy[i];
            if(dis[tx][ty] >= 0) {
                continue;
            }
            dis[tx][ty] = dis[op.x][op.y] + 1;
            Q.push(Sta(tx, ty));
        }
    }

    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= m; j++) {
            printf("%d ", dis[i][j]);
        }
        if(i < n) {
            puts("");
        }
    }

    return 0;
}

代码里的G[N][N]应该是多余的...