POJ 3984 迷宫问题

题意：输入一个5*5的迷宫地图，输出一条最短路径。

思路：BFS+路径输出。裸的路径输出问题。开一个存路径的结构体，每个点存来的点的坐标，然后递归到起点，回溯输出路径。

//#include<bits/stdc++.h>#include<iostream>#include<cstdio>#include<queue>using namespace std;int n, m, MAP[15][15], dir[4][2] = {{1, 0}, {0, 1}, {0, -1}, {-1, 0}};bool vis[15][15];pair<int, int> path[15][15], NOW;void bfs(){    queue<pair<int, int> > q;    q.push(make_pair(0, 0));    while (!q.empty())    {        NOW = q.front();        q.pop();        if (NOW.first == 4 && NOW.second == 4) return ;        for (int i = 0; i < 4; i++)        {            int X = NOW.first + dir[i][0], Y = NOW.second + dir[i][1];            if (X >= 0 && Y >= 0 && X < n && Y < m && MAP[X][Y] == 0 && !vis[X][Y])            {                q.push(make_pair(X, Y));                path[X][Y].first = NOW.first, path[X][Y].second = NOW.second;            }        }    }}void putout(int x, int y){    if (x == 0 && y ==0)    {        printf("(0, 0)\n");        return ;    }    putout(path[x][y].first, path[x][y].second);    printf("(%d, %d)\n",x ,y);}int main(){    n = 5; m = 5;    for (int i = 0; i < n; i++)    {        for (int j = 0; j < m; j++)        {            scanf("%d",&MAP[i][j]);        }    }    bfs();    putout(4, 4);    return 0;}/*0 1 0 0 00 1 0 1 00 0 0 0 00 1 1 1 00 0 0 1 0*/