HDU 1242 Rescue 优先队列+BFS

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Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12171    Accepted Submission(s): 4435

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

 

 

Source

ZOJ Monthly, October 2003

 

 

Recommend

Eddy

 

 

题目的意思是说friends用最短的时间去救angel， '.'表示通道 ，'#'表示墙壁， 'x'表示guard走一格要一单位时间，杀死一个guard要一个单位时间，求最短时间。WA了20来次才过了，我改了改发现到最后AC的一次仅仅是把SCANF改成CIN。哦，shit，太坑爹了。。。。。

 

#include
#include
#include
#include
using namespace std;
char map[200][200];
int dir[4][2]= {{0,1},{1,0},{0,-1},{-1,0}};
int vis[200][200];
int n,m,sx,sy,ans;
struct node
{
    int x,y,step;
     bool operator<(const node t)const
    {
        return t.step=1&&x<=n&&y>=1&&y<=m&&map[x][y]!='#'&&!vis[x][y])
        return true;
    return false;
}
node qq;
void BFS(int x,int y)
{
    priority_queueq;
    node next;
    qq.x=x;
    qq.y=y;
    qq.step=0;
    q.push(qq);
    vis[qq.x][qq.y]=1;
    while(!q.empty())
    {
        node first;
        first=q.top();
        q.pop();
        for(int i=0; i<4; i++)
        {
            int xx,yy;
            xx=first.x+dir[i][0];
            yy=first.y+dir[i][1];

            if(xx>=0&&xx=0&&yy>n>>m)
    {

        for(int i=0; i>map[i][j];
                if(map[i][j]=='a')
                {
                    sx=i;
                    sy=j;
                }
            }
            //getchar();
        }
        ans=-1;
        memset(vis,0,sizeof(vis));
        BFS(sx,sy);
        if(ans!=-1)
            printf("%d\n",ans);
        else
            printf("Poor ANGEL has to stay in the prison all his life.\n");
    }
    return 0;
}