《啊哈！演算法之解救小哈》

我們需要用dfs()函式處理的問題是：先檢查小哼是否已經到達小哈的位置，如果沒有到達則找出下一步可以走的地方

1. 判斷是否已經到達小哈的位置

    if((x == endx) && (y == endy))
    {
        if(step < minute)
            minute = step;
        return ;
    }

1. 找到下一步可以走的地方

int next[4][2]={{1,0},
                    {0,1},
                    {0
 
,-1},
                    {-1,0}};
for(int k=0;k<4;k++)
    {
        int kx=x+next[k][0];
        int ky=y+next[k][1];
        if((kx < 1) || (ky <1) || (kx > row) || (ky > col))
            continue;
        if((a[kx][ky] == 0) && (book[kx][ky] == 0))
        {
            book[kx][ky]=1
 
;
            dfs(kx,ky,step+1);
            book[kx][ky]=0;
        }
    }
    return ;

故整個程式為：

#include<iostream>
using namespace std;
int row=0,col=0,minute=99999;
int a[51][51],book[51][51];
int starty,startx=0;
int endx,endy=0;

void dfs(int x,int y,int step)
{
    int next[4][2]={{1,0},
                    {0
 
,1},
                    {0,-1},
                    {-1,0}};
    if((x == endx) && (y == endy))
    {
        if(step < minute)
            minute = step;
        return ;
    }
    //four ways
    for(int k=0;k<4;k++)
    {
        int kx=x+next[k][0];
        int ky=y+next[k][1];
        if((kx < 1) || (ky <1) || (kx > row) || (ky > col))
            continue;
        if((a[kx][ky] == 0) && (book[kx][ky] == 0))
        {
            book[kx][ky]=1;
            dfs(kx,ky,step+1);
            book[kx][ky]=0;
        }
    }
    return ;
}
int main()
{
    cout << "please input the row of the road: " ;
    cin >> row;
    cout << "please input the col of the road: ";
    cin >>col;
    for(int i=1;i<=row;i++)
    {
        for(int j=1;j<=col;j++)
            cin >> a[i][j];
    }
    int sx,sy,ex,ey=0;
    cout <<"please input the start x,y: " ;
    cin >> startx >> starty;
    cout << "please input the end x,y: ";
    cin >> endx>> endy;
    for(int i=1;i<=row;i++)
    {
       for(int j=1;j<=col;j++)
       {
           cout << a[i][j];
       }
       cout << endl;
    }
    book[startx][starty]=1;
    dfs(startx,starty,0);
    cout << minute;
    return 0;
}