Fire! UVA

链接

直接BFS搜索即可
需要注意的是每一轮要先搜索火然后再搜索人走的位置

#include <bits/stdc++.h>
using namespace std;

constexpr int N = 0x3f3f3f3f;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    // freopen("in.dat", "r", stdin);
    int t;
    cin >> t;
    while (t--)
    {
        int n, m;
        cin >> n >> m;
        vector<string> maze(n);
        for (int i = 0; i < n; i++)
            cin >> maze[i];
        // unordered_set<int> used;
        auto get = [](int x, int y) {
            return x * 10000 + y;
        };
        auto valid = [&](int x, int y) {
            return x <= 0 || x > n || y <= 0 || y > m || (maze[x - 1][y - 1] != '#' && maze[x - 1][y-1]!= 'F'&&maze[x-1][y-1]!='J'); //在这里面但是没有超出去
        };
        auto out = [&](int x, int y) {
            return x <= 0 || x > n || y <= 0 || y > m;
        };
        queue<pair<int, int>> que;
        int ans = 0x3f3f3f3f;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                if (maze[i][j] == 'F')
                    que.push({get(i + 1, j + 1), -100000000}); //火表示的是负数
            }
        }
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                if (maze[i][j] == 'J')
                    que.push({get(i + 1, j + 1), 0});
            }
        }
        //保证先搜了火
        int x, y;
        int dir[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        while (!que.empty())
        {
            auto e = que.front();
            que.pop();
            if (e.second >= 0 && out(e.first / 10000, e.first % 10000))
            {
                ans = e.second;
                break;
            }
            if (out(e.first / 10000, e.first % 10000))
                continue;
            for (int i = 0; i < 4; i++)
            {
                x = e.first / 10000 + dir[i][0];
                y = e.first % 10000 + dir[i][1];
                if (valid(x, y)){
                    que.push({get(x, y), e.second + 1});
                    if (!out(x, y))
                        maze[x - 1][y - 1] = maze[e.first / 10000-1][e.first % 10000-1];
                }
                
            }
        }
        if (ans == N)
            cout << "IMPOSSIBLE" << endl;
        else
            cout << ans << endl;
    }
}