FZU

题意，10*10的地图，有若干块草地“#”，草地可以点燃，并在一秒后点燃相邻的草地。有墙壁‘·‘阻挡。初始可以从任意两点点火。问烧完最短的时间。若烧不完输出-1.

题解：由于100的数据量，直接暴力。枚举两个起点，推入队列，然后bfs。烧完就返回深度，更新一个min值。

坑：（帮同学照bug）

return t.step+1;bfs后没有算上最后一步

ac代码

#define  _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string.h>
#include <queue>
#include <vector>
using namespace std;
char map[15][15];
int vis[15][15];
int T, n, m, ans, num;
struct Node {
    int i, j, step;
    Node(int i = 0, int j = 0, int step = 0) :i(i), j(j), step(step) {}
};
int dir[4][2] = { { 1,0 },{ -1,0 },{ 0,1 },{ 0,-1 } };
bool check(int i, int j) {
    if (i<0 || i >= n || j<0 || j >= m || vis[i][j] || map[i][j] == '.') return true;
    return false;
}
int bfs(int i, int j, int x, int y) {
    queue<Node>q;
    q.push(Node(i, j, 0)); q.push(Node(x, y, 0));
    vis[i][j] = 1; vis[x][y] = 1;
    int cnt;
    if (i == x&&j == y) cnt = 1;
    else cnt = 2;
    while (!q.empty()) {
        Node t = q.front();
        q.pop();
        for (int k = 0; k<4; k++) {
            int di = t.i + dir[k][0];
            int dj = t.j + dir[k][1];
            if (check(di, dj)) continue;
            q.push(Node(di, dj, t.step + 1));
            vis[di][dj] = 1;
            cnt++;
        }
        if (cnt >= num) { 
            //cout << i << ' ' << j << ' ' << x << ' ' << y <<' '<<t.step<< endl;
            return t.step+1; }
    }
    return -1;
}
int main() {
    scanf("%d", &T);
    for (int t = 1; t <= T; t++) {
        ans = (int)1e9; num = 0;
        scanf("%d%d", &n, &m);
        for (int i = 0; i<n; i++) {
            scanf("%s", map[i]);
            for (int j = 0; j<m; j++) if (map[i][j] == '#') num++;
        }
        if (num <= 2) { printf("Case %d: %d
", t, 0); continue; }
        for (int i = 0; i<n; i++) {
            for (int j = 0; j<m; j++) {
                if (map[i][j] != '#') continue;
                for (int x = 0; x<n; x++) {
                    for (int y = 0; y<m; y++) {
                        if (map[x][y] != '#') continue;
                        memset(vis, 0, sizeof(vis));
                        int tmp = bfs(i, j, x, y);
                        if (tmp>0) ans = min(ans, tmp);
                    }
                }
            }
        }
        if (ans == (int)1e9) ans = -1;
        printf("Case %d: %d
", t, ans);
    }
}