Time Limit: 1000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u

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

 

Source

ZOJ Monthly, October 2003


思路:单起点多终点求最短路,用BFS。由于走到有守卫的位置时间开销为2,需要使用优先队列,花费时间小的先出队。剩下的就是裸BFS。

AC代码:

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
int n, m;
char map[210][210];
bool vis[210][210];
const int dx[] = {0, 0, 1, -1},
			dy[] = {1, -1, 0, 0};
struct node
{
	int x, y, step;
	bool operator < (const node b) const
	{
		return step > b.step;
	}
}start;
priority_queue <node> q;
bool input()
{
	if (scanf("%d %d\n", &n, &m) == 2)
	{
		for (int i = 0; i < n; i++)
			scanf("%s\n", map[i]);
		return 1;
	}
	return 0;
}
void solve()
{
	for (int i = 0; i < n; i++)
		for (int j = 0; j < m; j++)
			if (map[i][j] == 'a')
				start = (node){i, j, 0};
	memset(vis, 0, sizeof(vis));
	while (!q.empty())
		q.pop();
	q.push(start);
	while (!q.empty())
	{
		node u = q.top();
		q.pop();
		if (vis[u.x][u.y])
			continue;
		vis[u.x][u.y] = 1;
		if (map[u.x][u.y] == 'r')
		{
			printf("%d\n", u.step);
			return;
		}
		node next;
		for (int i = 0; i < 4; i++)
		{
			next.x = u.x + dx[i];
			next.y = u.y + dy[i];
			if (next.x < 0 || next.x >= n || next.y < 0 || next.y >= m || map[next.x][next.y] == '#' || vis[next.x][next.y])
				continue;
			if (map[next.x][next.y] == 'x')
				next.step = u.step + 2;
			else
				next.step = u.step + 1;
			q.push(next);
		}
	}
	printf("Poor ANGEL has to stay in the prison all his life.\n");
}
int main()
{
	while (input())
		solve();
	return 0;
}

 

分类: ACM/ICPC

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