Rescue（BFS）

Rescue

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 9 Accepted Submission(s) : 4

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

Author

CHEN, Xue

题目大意：输入N行M列的字符串

          a代表所要查找的目标，r为起点，每移动一步则耗时1，遇到x耗时2，#为不可移动点

          问找到a时所用最少时间

  

题解：bfs即可，可能有多个’r’，而’a’只有一个，从’a’开始搜，找到的第一个’r’即为所求

需要注意的是这题宽搜时存在障碍物，遇到’x’点是，时间+2，如果用普通的队列就并不能保证每次出队的是时间最小的元素，所以要用优先队列

优先队列(priority_queue)的基本操作：

empty(); 队列为空返回1

pop(); 删除队首值

push(); 将元素归入

top(); 返回队列中优先级最高的元素

size(); 返回队列中元素的个数

具体见：http://blog.csdn.net/flqbestboy/article/details/7830782

#include
#include
using namespace std;
char map[210][210];//路线输入
bool mark[210][210];//标记此节点是否走过
int move[4][2]={{0,1},{1,0},{0,-1},{-1,0}};//上下左右移动
int m,n,i,j,res;
struct node
{
    int x,y,t;
    friend bool operator<(const node &n1,const node &n2)
    {
        return n1.t>n2.t;
    }//operator重载排序使，排序从大到小，队首值为最小值
}a,p,q;
int main()
{
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        for(i=0;iQ;
        Q.push(a);
        while(!Q.empty())
        {
            q=Q.top(),Q.pop();
            if(map[q.x][q.y]=='r')
            {
                res=q.t;
                break;
            }
            for(i=0;i<4;i++)
            {
                p.x=q.x+move[i][0],p.y=q.y+move[i][1];
                if(p.x>=0&&p.x=0&&p.y