POJ 3984 迷宫问题(bfs+pair)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
pair<int,int> Path[6][6],Now,Next;        // Path数组用来标记路径
int dir[4][2] = {1,0,0,1,-1,0,0,-1};
bool vis[6][6];
int MAP[6][6];
int step;

void bfs(){
  step = 0;
  memset(vis,0,sizeof(vis));
  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++){
      Next.first = Now.first + dir[i][0];
      Next.second = Now.second + dir[i][1];
     if(Next.first>=0&&Next.second>=0&&Next.first<5&&Next.second<5&&MAP[Next.first][Next.second]==0&&!vis[Next.first][Next.second]){
        vis[Next.first][Next.second] = 1;
        Path[Next.first][Next.second].first = Now.first;            // 将Path数组的Next保存为上一个点的坐标Now
        Path[Next.first][Next.second].second = Now.second;
        q.push(make_pair(Next.first,Next.second));
      }
    }
  }
}

void output(int x,int y){               // 递归过程
  if(x == 0 && y == 0){
    printf("(%d, %d)\n",x,y);
    return ;
  }
  output(Path[x][y].first,Path[x][y].second);
  printf("(%d, %d)\n",x,y);
}

int main()
{
  for(int i=0;i<5;i++){
    for(int j=0;j<5;j++){
      scanf("%d",&MAP[i][j]);
    }
  }
  bfs();
  output(4,4);
  return 0;
}