zoukankan      html  css  js  c++  java
  • HDU 1242 Rescue 营救天使

    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
     
      1 #include<cstdio>
      2 #include<string.h>
      3 #include<queue>
      4 using namespace std;
      5 int m,n,i,j,ans,map[205][205],ex,ey,bx,by;
      6 int dx[4]={-1,1,0,0};
      7 int dy[4]={0,0,-1,1};
      8 char str[205];
      9 struct stu
     10 {
     11     int x,y,step;
     12     bool friend operator <(stu a,stu b)
     13     {
     14         return a.step>b.step;
     15     }
     16 }st;
     17 int bfs()
     18 {
     19     priority_queue<stu>que;
     20     int x,y,i,time;
     21     stu next;
     22     
     23     st.x=bx;
     24     st.y=by;
     25     st.step=0;
     26     que.push(st);
     27     while(!que.empty())
     28     {
     29         
     30         st=que.top();
     31         que.pop();
     32         time=st.step;
     33         x=st.x;
     34         y=st.y;
     35         for(i=0;i<4;i++)
     36         {
     37             st.x=x+dx[i];
     38             st.y=y+dy[i];
     39             if(st.x>=0 &&st.y>=0&&st.x<m&&st.y<m&&map[st.x][st.y]!=0)
     40             {
     41                 if(st.x == ex && st.y == ey)
     42                 {
     43                     return time+1;
     44                 }
     45                 if(map[st.x][st.y] == 2)
     46                 {
     47                     st.step=time+2;
     48                 }
     49                 else 
     50                 {
     51                     st.step=time+1;
     52                 }
     53                 map[st.x][st.y]=0;
     54                 que.push(st);
     55             }
     56         }
     57     }
     58     return 0;
     59 }
     60 int main()
     61 {
     62     while(scanf("%d %d",&m,&n)!=EOF)
     63     {
     64         memset(map,0,sizeof(map));
     65         for(i = 0 ; i < m ; i++)
     66         {
     67             scanf("%s",&str);
     68             for(j = 0 ;j < n ; j++)
     69             {
     70                 if(str[j] == '.')
     71                 {
     72                     map[i][j]=1;
     73                 }
     74                 if(str[j] == 'x')
     75                 {
     76                     map[i][j]=2;
     77                 }
     78                 if(str[j] == 'a')
     79                 {
     80                     map[i][j]=1;
     81                     ex=i;
     82                     ey=j;
     83                 }
     84                 if(str[j] == 'r')
     85                 {
     86                     map[i][j]=1;
     87                     bx=i;
     88                     by=j;
     89                 }
     90             }
     91         }    
     92     /*    for( i = 0 ; i < m ; i++)
     93          {
     94             for(j =0 ; j < n ; j++)
     95              {
     96                 printf("%d  ",map[i][j]);
     97                  if(j == n-1)
     98                  {
     99                      printf("
    ");
    100                  }
    101             }
    102          }            */
    103         map[bx][by]=0;
    104         ans=bfs();
    105         if(ans)
    106             printf("%d
    ",ans);
    107         else
    108             printf("Poor ANGEL has to stay in the prison all his life.
    ");
    109         
    110     }
    111 }
    ——将来的你会感谢现在努力的自己。
  • 相关阅读:
    [转] 微服务架构
    vue拦截器实现统一token,并兼容IE9验证
    一句话 道出设计模式
    老话闲说,关于身份证号码的验证
    值得珍藏的HTTP协议详解
    微信内嵌H5网页 解决js倒计时失效
    使用JavaScript重定向URL参数
    Nodejs --我自己的学习笔记
    ONLY三行脚本 SQL数据恢复到指定时间点
    [转] 数据库加锁 sql加锁的
  • 原文地址:https://www.cnblogs.com/yexiaozi/p/5719387.html
Copyright © 2011-2022 走看看