HDU1010——Tempter of the Bone【DFS，奇偶剪枝，路径剪枝】

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分类专栏： DFS 文章标签： 2018暑假集训

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Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 146756 Accepted Submission(s): 39139

Problem Description

The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze began to shake, and the doggie could feel the ground sinking. He realized that the bone was a trap, and he tried desperately to get out of this maze.

The maze was a rectangle with sizes N by M. There was a door in the maze. At the beginning, the door was closed and it would open at the T-th second for a short period of time (less than 1 second). Therefore the doggie had to arrive at the door on exactly the T-th second. In every second, he could move one block to one of the upper, lower, left and right neighboring blocks. Once he entered a block, the ground of this block would start to sink and disappear in the next second. He could not stay at one block for more than one second, nor could he move into a visited block. Can the poor doggie survive? Please help him.

Input

The input consists of multiple test cases. The first line of each test case contains three integers N, M, and T (1 < N, M < 7; 0 < T < 50), which denote the sizes of the maze and the time at which the door will open, respectively. The next N lines give the maze layout, with each line containing M characters. A character is one of the following:

'X': a block of wall, which the doggie cannot enter;
'S': the start point of the doggie;
'D': the Door; or
'.': an empty block.

The input is terminated with three 0's. This test case is not to be processed.

Output

For each test case, print in one line "YES" if the doggie can survive, or "NO" otherwise.

Sample Input

4 4 5 S.X. ..X. ..XD .... 3 4 5 S.X. ..X. ...D 0 0 0

Sample Output

NO YES

Author

ZHANG, Zheng

Source

ZJCPC2004

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题目大意：从s点出发到达D点，问能不能恰好在T秒内到达。

这道题看似是DFS板子题，但同时要注意奇偶剪枝，否则会超时，关于奇偶剪枝可以参考一下這篇博客：https://blog.youkuaiyun.com/chyshnu/article/details/6171758。

同时可以借助这道题仔细了解一下DFS中的回溯过程。

#include<algorithm>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
using namespace std;
int n,m,t;
int sx,sy,ex,ey;
bool flag;
char mp[10][10];
bool vis[10][10];
int dx[4]={1,0,-1,0},dy[4]={0,-1,0,1};
void DFS(int x,int y,int ans){
    vis[x][y]=1;
    if(mp[x][y]=='D'&&ans==t){
        flag=1;
        return ;
    }
    if(ans>t) return;
    //奇偶剪枝
    int temp=t-ans-abs(ex-x)-abs(ey-y);//当前点到终点的距离
    if((temp&1)||temp<0) return ;
    for(int i=0;i<4;i++){
        int nx=x+dx[i];
        int ny=y+dy[i];
        if(nx>=0&&nx<n&&ny>=0&&ny<m&&!vis[nx][ny]&&mp[nx][ny]!='X'){
            //printf("(%d,%d)---\n",nx,ny);
            DFS(nx,ny,ans+1);
            if(flag) return ;
            vis[nx][ny]=0;//回溯
        }
    }
}
int main(){
    while(~scanf("%d%d%d",&n,&m,&t)&&n,m,t){
        memset(vis,0,sizeof(vis));
        flag=0;
        int sum=0;
        for(int i=0;i<n;i++) scanf("%s",mp[i]);
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(mp[i][j]=='S'){ sx=i;sy=j;}
                if(mp[i][j]=='D'){ ex=i;ey=j;}
                if(mp[i][j]=='X') sum++;
            }
        }
        //路径剪枝
        if((n*m-sum)>=t){
            DFS(sx,sy,0);
        }
        if(flag) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}