HDU1242:Rescue(BFS+优先队列)

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

 

 

题意：X代表卫兵，a代表终点，r代表起始点，.代表路，#代表墙

路花费一秒，x花费两秒

问到达终点的最少时间

思路：BFS+优先队列的果题

 

#include 
#include 
#include 
using namespace std;

struct node
{
    int x,y,step;
    friend bool operator<(node n1,node n2)
    {
        return n2.step=n || y>=m || !vis[x][y] || map[x][y] == '#')
        return 1;
    return 0;
}

int bfs()
{
    int i;
    priority_queue Q;
    node a,next;
    a.x = x1;
    a.y = y1;
    a.step = 0;
    Q.push(a);
    vis[x1][y1] = 0;
    while(!Q.empty())
    {
        a = Q.top();
        Q.pop();
        if(a.x == x2 && a.y == y2)
            return a.step;
        for(i = 0; i<4; i++)
        {
            next = a;
            next.x+=to[i][0];
            next.y+=to[i][1];
            if(check(next.x,next.y))//判断
                continue;
            next.step++;
            if(map[next.x][next.y] == 'x')//卫兵处多花费了一秒
                next.step++;
            if(vis[next.x][next.y]>=next.step)//存入最小时间
            {
                vis[next.x][next.y] = next.step;
                Q.push(next);
            }
        }
    }
    return 0;
}

int main()
{
    int i,j;
    while(~scanf("%d%d",&n,&m))
    {
        for(i = 0; i