地下迷宫（bfs输出路径）

题解：开一个pre数组用编号代替当前位置，编号用结构题另存，其实也可以i*m+j来代替，我写的有点麻烦了;

代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std;
#pragma(1)
typedef struct Node{
    int x, y;
    int t;
    int sz;
    friend bool operator < (Node a, Node b){
        return a.t > b.t;
    }
}Node;

Node node[];
priority_queue<Node>Q;
int vis[][];
int mp[][];
int pre[];
int disx[] = {,,,-};
int disy[] = {,-,,};
int mv(int x, int y){
    if(x == ){
        if(y == ){
            return ;
        }
        else if(y == -){
            return ;
        }
    }else if(x == ){
        if(y == )
            return ;
    }else if(x == -){
        if(y == )
            return ;
    }
}
void print(int sz){
    if(sz == )return;
    print(pre[sz]);
    printf(",[%d,%d]", node[sz].x, node[sz].y);
}
void bfs(int n, int m, int p){
    Node a,b;
    a.x = , a.y = , a.t = p, a.sz = ;
    while(!Q.empty()){
        Q.pop();
    }
    Q.push(a);
    memset(vis, , sizeof(vis));
    memset(pre, , sizeof(pre));
    memset(node, , sizeof(node));
    vis[][] = ;
    node[] = {, , , };
    int sz = ;
    while(!Q.empty()){
        a = Q.top();
        Q.pop();
        for(int i = ; i < ; i ++){
            b.x = a.x + disx[i];
            b.y = a.y + disy[i];
            b.t = a.t - mv(disx[i], disy[i]);
            b.sz = ++sz;
            node[sz] = {b.x, b.y, b.t, b.sz};
            pre[sz] = a.sz;
            if(b.x <  || b.y <  || b.x >= n || b.y >= m)continue;
            if(vis[b.x][b.y])continue;
            if(b.t < )continue;
            if(mp[b.x][b.y] == )continue;
            if(b.x ==  && b.y == m - ){
                printf("[%d,%d]",,);
                print(sz);
                puts("");
                return;
            }
            vis[b.x][b.y] = ;
            Q.push(b);
        }
    }
    puts("Can not escape!");
    return;
}
int main(){
    int n, m, q;
    while(~scanf("%d%d%d", &n, &m, &q)){
        memset(mp, , sizeof(mp));
        for(int i = ; i < n; i++){
            for(int j = ; j < m; j++){
                scanf("%d", &mp[i][j]);
            }
        }
        bfs(n, m, q);
    }
    return ;
}