迷宫2(bfs + 堆优化)

/**\
https://ac.nowcoder.com/acm/contest/26077/1005
从第i列第n行, 第1行m列添加起点, 然后bfs寻找能够分割两个点的墙

注意：
1、方向只能上下左右，因为对角线仍然可以穿过
2、处理图的数据的时候可以将-1的墙标记为成本为0，
   为0的则不能变为墙，标记为1
3、使用优先队列优化搜索的时间
\**/
#include <bits/stdc++.h>
#define int long long
using namespace std;

const int N = 1e3 + 10;
const int d[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

int n, m, t;
int dis[N][N], g[N][N];

struct node {
    int x, y, w;
    bool operator < (const node& rhs) const {
        return w > rhs.w;
    }
};

void accept() {
    for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= m; ++j) {
            dis[i][j] = 1e18;
            cin >> g[i][j];
            if(g[i][j] == -1) g[i][j] = 0;
            else if(g[i][j] == 0) g[i][j] = -1;
        }
    }

    priority_queue<node> q;

    for(int i = 1; i <= n; ++i) {
        if(g[i][1] != -1) {
            dis[i][1] = g[i][1];
            q.push({i, 1, g[i][1]});
        }
    }

    for(int i = 1; i <= m; ++i) {
        if(g[n][i] != -1) {
            dis[n][i] = g[n][i];
            q.push({n, i, g[n][i]});
        }
    }

    int ok = -1;

    while(!q.empty()) {
        auto now = q.top();
        q.pop();

        if(now.x == 1 || now.y == m) {
            ok = now.w;
            break;
        }

        for(int i = 0; i < 4; ++i) {
            int nx = now.x + d[i][0];
            int ny = now.y + d[i][1];

            if(nx < 1 || nx > n || ny < 1 || ny > m || g[nx][ny] == -1 || g[nx][ny] + now.w >= dis[nx][ny]) continue;

            dis[nx][ny] = g[nx][ny] + now.w;
            q.push({nx, ny, dis[nx][ny]});
        }
    }

    cout << ok << "\n";
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    cin >> t >> n >> m;
    for(int i = 0; i < t; ++i)
    {
        accept();
    }
    return 0;
}