码迷,mamicode.com
首页 > 其他好文 > 详细

Rescue

时间:2014-12-16 00:53:57      阅读:207      评论:0      收藏:0      [点我收藏+]

标签:des   blog   io   ar   os   sp   for   strong   on   

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
 
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<set>
#include<queue>
#include<vector>
using namespace std;
queue<int> q;
#define INF 1<<30
int n,m,sx,sy,ans,dp[205][205];
char a[205][205];
int dic[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
bool vis[205][205];
bool check(int x,int y)
{
      if(x<0||x>=n||y<0||y>=m||vis[x][y]||a[x][y]==‘#‘)
            return false;
      return true;
}
void bfs()
{
      int x,y,step;
      q.push(sx),q.push(sy),q.push(0);
      while(!q.empty())
      {
            x=q.front(),q.pop();
            y=q.front(),q.pop();
            step=q.front(),q.pop();
            if(a[x][y]==‘r‘)
            {
                  if(ans>step)
                        ans=step;
            }
            for(int i=0;i<4;i++)
            {
                  int xx,yy;
                  xx=x+dic[i][0],yy=y+dic[i][1];
                  if(check(xx,yy))
                  {
                       if((a[xx][yy]==‘.‘||a[xx][yy]==‘r‘)&&dp[xx][yy]>step+1)
                       {
                             q.push(xx),q.push(yy),q.push(step+1);
                             dp[xx][yy]=step+1;
                       }
                       if(a[xx][yy]==‘x‘&&dp[xx][yy]>step+2)
                       {
                             q.push(xx),q.push(yy),q.push(step+2);
                             dp[xx][yy]=step+2;
                       }
                  }
            }
      }
}
int main()
{
      while(scanf("%d%d",&n,&m)!=EOF)
      {
            ans=INF;
            memset(vis,0,sizeof(vis));
            for(int i=0;i<n;i++)
                  for(int j=0;j<m;j++)
                        dp[i][j]=INF;
            for(int i=0;i<n;i++)
                  scanf("%s",a[i]);
            for(int i=0;i<n;i++)
                  for(int j=0;j<m;j++)
                  {
                        if(a[i][j]==‘a‘)
                              sx=i,sy=j;
                  }
            vis[sx][sy]=1;
            bfs();
            if(ans!=INF)
                  printf("%d\n",ans);
            else
                  printf("Poor ANGEL has to stay in the prison all his life.\n");
      }
      return 0;
}

  

Rescue

标签:des   blog   io   ar   os   sp   for   strong   on   

原文地址:http://www.cnblogs.com/a972290869/p/4166201.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!