BFS PKU3984

题目已经说了有唯一的解。所以仅仅须要在找的过程中保存当前这个点前面的那个的点在队列中的位置

然后再输出的时候运用递归输出就能够了。

Description

int maze[5][5] = {

	0, 1, 0, 0, 0,

	0, 1, 0, 1, 0,

	0, 0, 0, 0, 0,

	0, 1, 1, 1, 0,

	0, 0, 0, 1, 0,

};

Input

Output

Sample Input

0 1 0 0 0
0 1 0 1 0
0 0 0 0 0
0 1 1 1 0
0 0 0 1 0

Sample Output

(0, 0)
(1, 0)
(2, 0)
(2, 1)
(2, 2)
(2, 3)
(2, 4)
(3, 4)
(4, 4)

#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <limits.h>
#include <ctype.h>
#include <string.h>
#include <string>
#include <queue>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <stack>
#include <deque>
#include <vector>
#include <set>
using namespace std;
#define MAXN 10

int xx[4]={-1,0,1,0};
int yy[4]={0,1,0,-1};
int vis[MAXN][MAXN];
int map[MAXN][MAXN];
int ans;

struct node{
    int x,y;
    int pre;
}q[111];
int start=0;
int end = 1;

void print(int x){
    if(q[x].pre != -1){
        print(q[x].pre);
        printf("(%d, %d)\n",q[x].x,q[x].y);
    }
}

void BFS(int x,int y){
    int i;
    vis[x][y] = 1;
    q[start].x = x;
    q[start].y = y;
    q[start].pre = -1;
    while(start < end){
        if(q[start].x==4 && q[start].y==4){
            print(start);
        }
        for(i=0;i<4;i++){
            int dx = xx[i] + q[start].x;
            int dy = yy[i] + q[start].y;
            if(dx>=0 && dx<5 && dy>=0 && dy<5 && vis[dx][dy]==0 && map[dx][dy]==0){
                q[end].x = dx;
                q[end].y = dy;
                vis[dx][dy] = 1;
                q[end].pre = start;
                end++;
            }
            //if(dx==4 && dy==4){
            //    print(start);
            //}
        }
        start++;
    }
}

int main(){
    int i,j;
    for(i=0;i<5;i++){
        for(j=0;j<5;j++){
            cin>>map[i][j];
        }
    }
    memset(vis,0,sizeof(vis));
    printf("(0, 0)\n");
    BFS(0,0);
    //printf("(4, 4)\n");

    return 0;
}