zoj 1649 Rescue bfs

Rescue

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

题意：Angel被困在一个图里面了，它的朋友要去救她，但是图中有些守卫，她朋友比较666，对于战斗力只有5的侍卫渣可以随便解决掉，但还是要花费一个单位的时间（也许是作为一名光荣的骑士为他剑下亡魂祈祷），朋友们也可能不在一个位置，她朋友找到她最少需要多久。

思路：找到她每一个朋友的位置，每个进行一次bfs，求出最小值即可。

代码：

#include <cstdio>#include <queue>#include <cstring>using namespace std;#define maxn 200+10struct Node{int x,y;int step;bool friend operator <(Node a,Node b){return a.step > b.step;}}start[maxn],stop;int dx[] = {0,0,-1,1};int dy[] = {1,-1,0,0};int n,m;char s[maxn][maxn];int ans;int cnt;bool bfs(int k){priority_queue<Node> que;bool v[maxn][maxn];memset(v,false,sizeof v);Node node;que.push(start[k]);v[start[k].x][start[k].y] = true;while(!que.empty()){node = que.top();que.pop();if(node.x==stop.x && node.y==stop.y){ans = node.step;return true;}for(int i=0;i<4;i++){int xx = node.x+dx[i], yy = node.y+dy[i];if(xx>=0 && xx<n && yy>=0 && yy<m && !v[xx][yy] && s[xx][yy]!='#'){Node temp = node;temp.x = xx, temp.y = yy;if(s[xx][yy]=='x') temp.step += 2;else temp.step++;que.push(temp);v[xx][yy] = true;}}}return false;}int main(){while(~scanf("%d %d\n",&n,&m)){cnt = 0;for(int i=0;i<n;i++){for(int j=0;j<m;j++){scanf("%c",&s[i][j]);if(s[i][j]=='r'){start[cnt].x = i;start[cnt].y = j;start[cnt++].step = 0;}if(s[i][j]=='a'){stop.x = i;stop.y = j;}}getchar();}bool flag = false;int res = 9999999;for(int i=0;i<cnt;i++){if(bfs(i)){flag = true;if(ans<res) res = ans;}}if(flag) printf("%d\n",res);else printf("Poor ANGEL has to stay in the prison all his life.\n");}return 0;}