广度优先遍历求解 ZOJ 649 （Rescue）

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

广度优先遍历求解。 不同于一般的迷宫最短路径求解，每步的时间是不确定的。因此需要用一个附加数据保存时间，并遍历所有可达的情况。

那么BFS是否会无限搜索下去呢？不会的，因为我们加入了判断条件是，这种走法比之前所花费的时间更少。

#include <iostream>
#include <queue>
#include <stdio.h>
using namespace std;
#define MAXMN 200
#define INF 1000000

class Point{
	public:
	int x,y,time;
};

queue<Point> Q;
int N,M;
char map[MAXMN][MAXMN];
int mintime[MAXMN][MAXMN]; //表示到达当前位置所需要的 最短时间
int dir[4][2] = { { -1, 0 }, { 0, -1 }, { 1, 0 }, { 0, 1 } };
int ax,ay; //结束位置

int bfs(Point s){ //从s点开始搜索

	Q.push(s);
	Point hd;
	while( !Q.empty()){
		hd = Q.front();
		Q.pop();
		for(int i=0; i<4; i++){
			int x = hd.x + dir[i][0];
			int y = hd.y + dir[i][1];
			if(x >= 0 && x <N && y >=0 && y < M && map[x][y] != '#'){
				Point t;
				t.x = x;
				t.y = y;
				t.time = hd.time + 1;
				if(map[x][y] == 'x') t.time ++;
				if(t.time < mintime[x][y]){
					mintime[x][y] = t.time;
					Q.push(t);
				}
			}
		}
	}
	return mintime[ax][ay];
}

int main() {

	while(scanf("%d %d", &N,&M) != EOF){
		for(int i=0; i<N; i++)
			scanf("%s", map[i]);

		int sx,sy;
		Point start;
		for(int i=0; i<N; i++){
			for(int j=0; j<M; j++){
				mintime[i][j] = INF; //mintime初始为最大，不可达
				if(map[i][j] == 'a'){
					ax = i;
					ay = j;
				}else if(map[i][j] == 'r'){
					sx = i;
					sy = j;
				}
			}
		}

		start.x = sx;
		start.y = sy;
		start.time = 0;
		mintime[sx][sy] = 0;

		int mint = bfs(start);
		if(mint < INF)
			printf("%d\n",mint);
		else
			printf("Poor ANGEL has to stay in the prison all his life.\n");
	}
	return 0;
}