HDOJ--1010题C++

有两点需要注意，第一点就是，题意是指在第T秒时门才打开，并不是越早到门口就能越早出去，而是只要存在指定长的路径即可以出去。

第二点，每次深搜发现答案不符都必须将点重新初始化。
#include<iostream>
#include<utility>
#define INF 10000
using namespace std;
char maze[100][7];
int d[100][7];
int n,m,t,start_i,start_j,end_i,end_j;
int dx[]= {0,1,0,-1},dy[]= {1,0,-1,0};
bool dfs(int i,int j);
int main()
{
    while(cin>>n>>m>>t && n &&m
            && t)
    {
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
            {
                cin>>maze[i][j];
                if(maze[i][j]==‘S‘)
                {
                    start_i=i;
                    start_j=j;
                }
                if(maze[i][j]==‘D‘)
                {
                    end_i=i;
                    end_j=j;
                }
            }
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
                d[i][j]=INF;
        d[start_i][start_j]=0;
        if(dfs(start_i,start_j))
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return 0;
}
bool dfs(int i,int j)
{
    if(i==end_i && j==end_j)
    {
        if(d[i][j]==t)
        {
            d[i][j]=INF;
            return true;
        }
        else
        {
            d[i][j]=INF;
            return false;
        }
    }
    else
    {
        for(int k=0; k<4; k++)
        {
            int nx=dx[k]+i,ny=dy[k]+j;
            if( nx>=0 &&nx < n &&ny>=0 && ny<m && (d[nx][ny]==INF) &&((maze[nx][ny]==‘.‘)||(maze[nx][ny]==‘D‘)))
            {

                    d[nx][ny]=d[i][j]+1;
                    if(dfs(nx,ny))
                    {
                        d[nx][ny]=INF;
                        return true;
                    }
                    else
                        d[nx][ny]=INF;
            }
        }
        d[i][j]=INF;
        return false;
    }
}