poj 3984 迷宫问题 BFS

/*

题目：

       求最少时间从(0,0)走到(4,4)的路径

分析：

       纯粹BFS题目，不过需要打印路径，可以用数组记录当前的坐标的上一个坐标，

       因为BFS构造出一棵BFS最优生成树，每一个节点的父母节点都是唯一的，具体

       参考算法导论。。。

*/

#include <iostream>

#include <cstring>

#include <cstdio>

#include <queue>

using namespace std;

#define X 6

int map[X][X],pre[X][X];

bool visit[X][X];

struct node

{

       int x,y;

};

void print(int x,int y)     //打印路径

{

       if(x||y)    //一直递归到起始的(0,0)为止，然后不停输出到(4,4)走过的路

              print(pre[x][y]/10,pre[x][y]%10);

       if(pre[x][y]!=-1&&pre[x][y]!=-1)       //不输出(0,0)以前的坐标。。。

              printf("(%d, %d)\n",pre[x][y]/10,pre[x][y]%10);

}

void bfs()

{

       queue<node> q;

       node temp,cur;

       temp.x = 0;

       temp.y = 0;

       q.push(temp);

       int x,y;

       visit[0][0] = true;

       while(!q.empty())   //BFS核心

       {

              cur = q.front();

              q.pop();

              x = cur.x;

              y = cur.y;

              if(x==4&&y==4)

                     print(4,4);

              if(x&&!visit[x-1][y]&&!map[x-1][y])//往上走

              {

                     visit[x-1][y] = true;

                     temp.x = x-1;

                     temp.y = y;

                     pre[x-1][y] = x*10+y;

                     q.push(temp);

              }

              if(y&&!map[x][y-1]&&!visit[x][y-1]) //往左走

              {

                     visit[x][y-1] = true;

                     temp.x = x;

                     temp.y = y-1;

                     pre[x][y-1] = x*10+y;

                     q.push(temp);

              }

              if(y<4&&!map[x][y+1]&&!visit[x][y+1])//往右走

              {

                     visit[x][y+1] = true;

                     temp.x = x;

                     temp.y = y+1;

                     pre[x][y+1] = x*10+y;

                     q.push(temp);

              }

              if(x<4&&!map[x+1][y]&&!visit[x+1][y])//往下走

              {

                     visit[x+1][y] = true;

                     temp.x = x+1;

                     temp.y = y;

                     pre[x+1][y] = x*10+y;

                     q.push(temp);

              }

       }

}

int main()

{

       freopen("sum.in","r",stdin);

       freopen("sum.out","w",stdout);

       while(cin>>map[0][0])

       {

              memset(pre,-1,sizeof(pre));

              memset(visit,false,sizeof(visit));

              for(int i=0;i<5;i++)

                     for(int j=0;j<5;j++)

                            if(i||j)

                                   scanf("%d",&map[i][j]);

              bfs();

              printf("(%d, %d)\n",4,4);//最后打印最后的坐标

       }

       return 0;

}