HDOJ-1242-Rescue

Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21644    Accepted Submission(s): 7721

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

13

 

 

 

第一個完整理解寫下來的bfs,曾經怎麼寫都不對的題,終於解決了.

題意:a 被抓走了, r 來救她(他), # 是牆, . 是路,走一步一分鐘, x 是守衛,遇見一個多花一分鐘,求最短需要多長時間

思路:基本是個模板題,不過按題意細節來看,friends來救她,有可能有多個 r ,遇見最近的一個就得救,因此逆向思維, a 只有一個,所以讓 a 去找最近的 r 即可(最後發現r去找a也會AC,想多了,不過嚴謹一些比較好).

#include<cstdio>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;
#define INF 0x3f3f3f3f
#define MAX 220
char map[MAX][MAX];
int d[4][2]{0,1,0,-1,1,0,-1,0};
int vis[MAX][MAX],N,M,sx,sy,ans;
struct node{
	int x,y,step;
};
bool operator <(const node &a,const node &b){
	if(a.step==b.step)
		return a.x<b.x;
	return b.step<a.step;
}
node now;
node then;
bool juge(int a,int b){
	if(a>N||a<1||b>M||b<1||vis[a][b]||map[a][b]=='#')
		return false;
	return true;
}
void bfs(){
	now.x=sx;
	now.y=sy;
	now.step=0;
	memset(vis,0,sizeof(vis));
	priority_queue<node>q;
	q.push(now);
	vis[sx][sy]=1;
	while(!q.empty()){
		now=q.top();
		q.pop();
		for(int v=0;v<4;v++){
			then.x=now.x+d[v][0];
			then.y=now.y+d[v][1];
			if(juge(then.x,then.y)){
				if(map[then.x][then.y]=='x')
					then.step=now.step+2;
				else
					then.step=now.step+1;
				if(map[then.x][then.y]=='r'){
					ans=then.step;
					return;
				}
				vis[then.x][then.y]=1;
				q.push(then);
			}
		}
	}
}
int main(){
	while(scanf("%d%d",&N,&M)!=EOF){
		for(int n=1;n<=N;n++){
			getchar();
			for(int m=1;m<=M;m++){
				scanf("%c",&map[n][m]);
				if(map[n][m]=='a'){
					sx=n;sy=m;//記錄下a的座標爲起點
				}
			}
		}
		ans=INF;
		bfs();
		if(ans!=INF)//ans值不變表示沒有可解救的通路
		printf("%d\n",ans);
		else
		puts("Poor ANGEL has to stay in the prison all his life.");
	}
	return 0;
}