TOJ 1335 HDU 1242 ZOJ 1649 营救天使 / 广搜+优先队列

营救天使

时间限制(普通/Java):1000MS/10000MS 运行内存限制:65536KByte



描述

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.)

输入

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, "x" stands for a guard, and "r" stands for each of Angel's friend.

Process to the end of the file.

输出

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."

样例输入

[code]7 8
#.#####.
#.a#..r.
#..#x...
..#..#.#
#...##..
.#......
........

样例输出

13

#include <stdio.h>
#include <string>
#include <iostream>
#include <queue>
#define MAX 210
#define inf 0x7fffffff
using namespace std;

char a[MAX][MAX];
int map[MAX][MAX];
int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};//代表上 下 左 右 
int n,m;
int Min;
struct point
{
	int x;
	int y;
	int step;//步数 
	friend bool operator <(point a,point b)
	{
		return a.step > b.step;//优先队列 时间少的先搜 
	}
}s,e;

void bfs()
{
	priority_queue <point> q;
	memset(map,0,sizeof(map));
	q.push(s);
	while(!q.empty())
	{
		point p = q.top();	
		q.pop();
		if(p.x == e.x && p.y == e.y)
		{
			if(p.step < Min)
				Min = p.step;
			return ;
		}
		for(int i = 0;i < 4; i++)
		{
			point t;
			t.x = p.x + dir[i][0];
			t.y = p.y + dir[i][1];
			if(t.x >= 0 && t.x < n && t.y >= 0 && t.y < m && a[t.x][t.y] != '#' && map[t.x][t.y] == 0)
			{
				t.step = p.step + 1;
				map[t.x][t.y] =  1;//标记走过了不再走  
				if(a[t.x][t.y] == 'x')
				{
					t.step++;
					a[t.x][t.y] = '.';//不管那个人 杀掉了就死了 这里wrong了一次 
				}				
				q.push(t);
			}
		}
	}
}
main()
{
	int i,j;
	while(scanf("%d %d %d",&n,&m)!=EOF)
	{
		Min = inf;
		for(i = 0;i < n; i++)
		{
			scanf("%s",a[i]);
			for(j = 0;j < m; j++)
			{
				if(a[i][j] == 'a')
				{
					e.x = i;
					e.y = j;
				}
			}
		}
		for(i = 0;i < n; i++)
		{
			for(j = 0;j < m; j++)
			{
				if(a[i][j] == 'r')
				{
					s.x = i;
					s.y = j;			
					bfs();				
				}
			}
		}
		if(Min != inf)
			printf("%d\n",Min);
		else
			puts("Poor ANGEL has to stay in the prison all his life.");
	}
}