HDU 1242 Rescue （深搜）

qdu_zhaiH

于 2019-03-10 20:23:44 发布

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分类专栏：日常刷题

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探讨了在一个由墙壁、道路和守卫构成的监狱中，寻找从天使位置到其朋友位置最短路径的算法。通过深度优先搜索，考虑了杀死守卫所需的时间，旨在找到最短的救援路径。

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Problem Description

Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison.

Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.

You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)

Input

First line contains two integers stand for N and M.

Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend.

Process to the end of the file.

Output

For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the prison all his life."

Sample Input

7 8 #.#####. #.a#..r. #..#x... ..#..#.# #...##.. .#...... ........

Sample Output

思路：就是简单的深搜，需要注意的就是可能有多个朋友，所以要从a 开始搜，然后更新Min

#include<iostream>
using namespace std;
int vis[205][205];
char m[205][205];
int N,M,x,y,Min;
void dfs(int x,int y,int len){
	if(x<0||x>=N||y<0||y>=M) return ;//越界 
	if(vis[x][y]==1)  return ;//已经标记过 
	if(len>=Min) return ;
	if(m[x][y]=='#') return ;
	if(m[x][y]=='r'){
		if(len<Min)
		    Min=len;
		return ;
	}
	if(m[x][y]=='x') len++;
	vis[x][y]=1;//标记 
	dfs(x+1,y,len+1);
	dfs(x-1,y,len+1);
	dfs(x,y+1,len+1);
	dfs(x,y-1,len+1);
	vis[x][y]=0;//取消标记 
}
int main(){
	while(scanf("%d%d",&N,&M)!=EOF){
		for(int i=0;i<N;i++){
		    for(int j=0;j<M;j++){
			    cin>>m[i][j];
			    if(m[i][j]=='a'){
				    x=i;
				    y=j;
		    	}//从a开始 
	    	}
		    getchar();
    	}
	    int len=0;
	    Min=99999;
	    dfs(x,y,len);
	    if(Min!=99999)
	        cout<<Min<<endl;
	    else
	        cout<<"Poor ANGEL has to stay in the prison all his life."<<endl;
    }
	return 0;
}