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)

解題思路

廣搜問題：在正常基礎上多定義一個父節點，可以根據結構體回溯

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int a[6][6],book[6][6];
struct node
{
    int x;
    int y;
    int s;
    int f;
}que[100];
int main()
{
    int next[4][2]={1,0,0,-1,-1,0,0,1};
    for(int i=0;i<5;i++)
    for(int j=0;j<5;j++)
    scanf("%d",&a[i][j]);
    int head=1,tail=1;
    que[head].x=0;
    que[head].y=0;
    que[head].s=0;
    que[head].f=0;
    book[1][1]=1;
    tail++;
    int flag=0,tx,ty;
    while(head<tail)
    {
        for(int k=0;k<=3;k++)
        {
            tx=que[head].x+next[k][0];
            ty=que[head].y+next[k][1];
            if(tx<0||tx>4||ty<0||ty>4)
            continue;
            if(a[tx][ty]==0&&book[tx][ty]==0)
            {
                book[tx][ty]=1;
                que[tail].x=tx;
                que[tail].y=ty;
                que[tail].s=que[head].s+1;
                que[tail].f=head;
                tail++;
            }
            if(tx==4&&ty==4)
            {
                flag=1;
                break;
            }

        }
        if(flag==1)
            break;
            head++;

    }
    int ax[30]={0},ay[30]={0};
    int z=-1;
    ax[++z]=que[tail-1].x;
    ay[z]=que[tail-1].y;
      int sum=que[tail-1].s;
   tail=tail-1;

    for(int i=1;i<sum;i++)
    {
        tail=que[tail].f;
        ax[++z]=que[tail].x;
        ay[z]=que[tail].y;
    }
    ax[++z]=0;
    ay[z]=0;

    for(int j=z;j>=0;j--)
    printf("(%d, %d)\n",ax[j],ay[j]);
    return 0;
}