zoukankan      html  css  js  c++  java
  • hdu1010 Tempter of the Bone(深搜+剪枝问题)

    Tempter of the Bone
    
    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 51193    Accepted Submission(s): 13775
    
    
    Problem Description
    The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze began to shake, and the doggie could feel the ground sinking. He realized that the bone was a trap, and he tried desperately to get out of this maze.
    
    The maze was a rectangle with sizes N by M. There was a door in the maze. At the beginning, the door was closed and it would open at the T-th second for a short period of time (less than 1 second). Therefore the doggie had to arrive at the door on exactly the T-th second. In every second, he could move one block to one of the upper, lower, left and right neighboring blocks. Once he entered a block, the ground of this block would start to sink and disappear in the next second. He could not stay at one block for more than one second, nor could he move into a visited block. Can the poor doggie survive? Please help him.
     
    
    Input
    The input consists of multiple test cases. The first line of each test case contains three integers N, M, and T (1 < N, M < 7; 0 < T < 50), which denote the sizes of the maze and the time at which the door will open, respectively. The next N lines give the maze layout, with each line containing M characters. A character is one of the following:
    
    'X': a block of wall, which the doggie cannot enter; 
    'S': the start point of the doggie; 
    'D': the Door; or
    '.': an empty block.
    
    The input is terminated with three 0's. This test case is not to be processed.
     
    
    Output
    For each test case, print in one line "YES" if the doggie can survive, or "NO" otherwise.
     
    
    Sample Input
    4 4 5
    S.X.
    ..X.
    ..XD
    ....
    3 4 5
    S.X.
    ..X.
    ...D
    0 0 0
     
    
    Sample Output
    NO
    YES
     
    
    Author
    ZHANG, Zheng
     
    
    Source
    ZJCPC2004
     
    
    Recommend
    JGShining

    #include <iostream>
    #include<cmath>
    #include<stdio.h>
    #include<stdlib.h>
    using namespace std;
    int sx,sy,ex,ey;
    int n,m;
    int flag;
    int d[4][2]={0,1,1,0,0,-1,-1,0};
    char map[10][10];
    void dfs(int x,int y,int t)
    {
        //cout<<x<<" "<<y<<" "<<t<<endl;
        if(flag==1)return;
        if(t<sqrt(float((ex-x)*(ex-x)+(ey-y)*(ey-y)))||(t-abs(ex-x)+abs(ey-y))%2)return;
        else if(t==0)
        {
            if(x==ex&&y==ey){flag=1;return;}
            else
            return;
        }
        else
        {
            for(int i=0;i<4;i++)
            {
                int nx = x+d[i][0];
                int ny = y+d[i][1];
                if(nx>0&&nx<=n&&ny>0&&ny<=m&&(map[nx][ny]=='.'||map[nx][ny]=='D'))
                {
                    map[nx][ny]='X';
                    dfs(nx,ny,t-1);
                    map[nx][ny]='.';
                }
            }
        }
        return ;
    }
    int main()
    {
        int t,num;
        char str[10];
        while(scanf("%d%d%d",&n,&m,&t),n||m||t)
        {
            num=0;
            for(int i=1;i<=n;i++)
            {
                scanf("%s",str);
                for(int j=1;j<=m;j++)
                {
                    map[i][j]=str[j-1];
                    if(map[i][j]=='S')sx=i,sy=j;
                    else if(map[i][j]=='D')ex=i,ey=j,num++;
                    else if(map[i][j]=='.')num++;
                }
            }
    
            flag=0;
            if(num>=t)
            dfs(sx,sy,t);
            if(flag==0)
            printf("NO
    ");
            else
            printf("YES
    ");
        }
        return 0;
    }
     
  • 相关阅读:
    多页面通信问题
    Web Worker
    http 状态码(转载)
    如何把一个Array 复制到ArrayList中?
    使用C#数据如何导到excel
    面试题:Web Service与wcf的区别
    ASP.NET为我们提供了几种错误处理机制?
    使用sql语句创建带有输出参数的存储过程
    存储过程维护
    Linux系统下的网络配置
  • 原文地址:https://www.cnblogs.com/heqinghui/p/3181831.html
Copyright © 2011-2022 走看看