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[10010];
priority_queue<Node>Q;
int vis[11][11];
int mp[11][11];
int pre[10010];
int disx[4] = {0,0,1,-1};
int disy[4] = {1,-1,0,0};
int mv(int x, int y){
    if(x == 0){
        if(y == 1){
            return 0;
        }
        else if(y == -1){
            return 3;
        }
    }else if(x == 1){
        if(y == 0)
            return 1;
    }else if(x == -1){
        if(y == 0)
            return 1;
    }
}
void print(int sz){
    if(sz == 0)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 = 0, a.y = 0, a.t = p, a.sz = 0;
    while(!Q.empty()){
        Q.pop();
    }
    Q.push(a);
    memset(vis, 0, sizeof(vis));
    memset(pre, 0, sizeof(pre));
    memset(node, 0, sizeof(node));
    vis[0][0] = 1;
    node[0] = {0, 0, 0, 0};
    int sz = 0;
    while(!Q.empty()){
        a = Q.top();
        Q.pop();
        for(int i = 0; i < 4; 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 < 0 || b.y < 0 || b.x >= n || b.y >= m)continue;
            if(vis[b.x][b.y])continue;    
            if(b.t < 0)continue;
            if(mp[b.x][b.y] == 0)continue;
            if(b.x == 0 && b.y == m - 1){
                printf("[%d,%d]",0,0);
                print(sz);
                puts("");
                return;
            }
            vis[b.x][b.y] = 1;
            Q.push(b);
        }
    }
    puts("Can not escape!");
    return;
}
int main(){
    int n, m, q;
    while(~scanf("%d%d%d", &n, &m, &q)){
        memset(mp, 0, sizeof(mp));
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                scanf("%d", &mp[i][j]);
            }
        }
        bfs(n, m, q);
    }
    return 0;
}