zoukankan      html  css  js  c++  java
  • HDU 1010 Temper of the bone(深搜+剪枝)

    Tempter of the Bone

    Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
    Total Submission(s) : 15   Accepted Submission(s) : 9

    Font: Times New Roman | Verdana | Georgia

    Font Size: ← →

    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
    #include <iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    using namespace std;
    int dr[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
    int f[10][10];
    char mp[10][10];
    int sx,sy,tx,ty,t,n,m,i,j;
    int check(int x,int y)
    {
        if (x>=0 && x<n && y>=0 && y<m && mp[x][y]!='X' && !f[x][y]) return 1;
          else return 0;
    }
    int  dfs(int x,int y,int time)
    {
        if (time==0 && x==tx && y==ty)
               return 1;
        for(int i=0;i<4;i++)
        {
            int xx=x+dr[i][0];
            int yy=y+dr[i][1];
            if (check(xx,yy))
            {
                f[xx][yy]=1;
                if (dfs(xx,yy,time-1)) return 1;
                f[xx][yy]=0;
            }
        }
        return 0;
    }
    
    int main()
    {
        while(scanf("%d%d%d",&n,&m,&t) && n!=0)
        {
            for(i=0;i<n;i++)
                {
                    scanf("%s",&mp[i]);
                    for(j=0;j<m;j++)
                        if (mp[i][j]=='S') sx=i,sy=j;
                           else if (mp[i][j]=='D') tx=i,ty=j;
                }
           if (t==0)
           {
                if (sx!=tx && sy!=ty) printf("NO\n");
                  else if (sx==tx && sy==ty) printf("YES\n");
                continue;
           }
           if (abs(tx-sx)+abs(ty-sy)>t || (t-abs(tx-sx)-abs(ty-sy))%2!=0) {printf("NO\n");continue;}
    
           memset(f,0,sizeof(f));
           f[sx][sy]=1;
           if (dfs(sx,sy,t)) printf("YES\n");
                    else printf("NO\n");
        }
    
        return 0;
    }
  • 相关阅读:
    javascript 事件冒泡
    Java 理论与实践: 正确使用 Volatile 变量
    Concurrency,Java 并发
    POJ2379 ACM Rank Table 模拟题
    HDU1711Number Sequence KMP
    POJ1061 青蛙的约会 扩展GCD
    HDU2523 SORT AGAIN HASH
    HDU2087剪花布条 KMP
    HDU3736 Cyclic Nacklace KMP
    HDU1709The Balance 母函数
  • 原文地址:https://www.cnblogs.com/stepping/p/5635867.html
Copyright © 2011-2022 走看看