hdu1242优先队列+BFS

2024年2月4日 195次阅读

Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29901    Accepted Submission(s): 10531

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     
简单的优先队列+BFS题，，本来是个普通的广搜过程，但是，在x也就是守卫处，时间+2，这样就会导致守卫周围的时间不同，所以，利用优先队列找到最短时间，继续BFS就可以找到最短时间的路了。
#include<stdio.h>
#include<string.h>
#include<queue>
#include<iostream>

#define MAX 205

using namespace std;

char map[MAX][MAX];
int visited[MAX][MAX];
int m,n;
int dir[4][2]={0,1,0,-1,1,0,-1,0};

typedef struct Node{
	int x;
	int y;
	int time;
}N;

struct cmp{
	bool operator ()(const N &a,const N &b)
	{
		if(a.time > b.time)
		{
			return true;
		}
		return false;
	}
};

bool able(int x,int y)
{
	if(x<0||y<0||x>m-1||y>n-1||map[x][y]=='#'||visited[x][y]!=0)
		return false;
	return true;
}

int bfs(int sx,int sy)
{
	priority_queue<Node,vector<Node>,cmp> q;
	Node a,b;
	a.x = sx;
	a.y = sy;
	a.time = 0;
	q.push(a);

	while(!q.empty())
	{
		a = q.top();
		q.pop();
		if(map[a.x][a.y]=='r')
			return a.time;
		for(int i=0;i<4;i++)
		{
			
			b.x = a.x+dir[i][0];
			b.y = a.y+dir[i][1];
			if(able(b.x,b.y))
			{
				if(map[b.x][b.y]=='x')
				{
					b.time = a.time+2;
				}
				else{
					b.time = a.time+1;
				}
				visited[b.x][b.y] = 1;
				q.push(b); 
			}
		}
	}
	return -1;
}

int main()
{
	
	int sx,sy;//设置起点座标 
	int i,j;
	while(~scanf("%d%d",&m,&n))
	{
		for(i=0;i<m;i++)
		{
			scanf("%s",map[i]);
			for(j=0;j<n;j++)
			{
				if(map[i][j]=='a')
				{
					sx = i;
					sy = j;
				}
			}
		}
		memset(visited,0,sizeof(visited));
		visited[sx][sy] = 1;
		int res = bfs(sx,sy);
		if(res==-1)  
        printf("Poor ANGEL has to stay in the prison all his life.\n");  
        else  
        printf("%d\n",res);
	}
	return 0;
}
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