zoukankan      html  css  js  c++  java
  • Tempter of the bone

    题目描述:

    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.

    输入:

    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.

    输出:

    For each test case, print in one line "YES" if the doggie can survive, or "NO" otherwise.

    样例输入:
    4 4 5
    S.X.
    ..X.
    ..XD
    ....
    3 4 5
    S.X.
    ..X.
    ...D
    0 0 0
    
    样例输出:
    NO
    YES
    
    提示:

     用scanf读取输入。

    #include<iostream>
    #include<stdio.h>
    #include<string>
    using namespace std;
    char maze[8][8];
    int go[8][2]={
    1,0,
    -1,0,
    0,1,
    0,-1,
    -1,-1,
    1,1,
    -1,1,
    1,-1
    };
    int m,n,t;
    bool suc;
    void dfs(int x,int y,int time){
        for (int i=0;i<4;i++){
            int nx=x+go[i][0];
            int ny=y+go[i][1];
            if (nx<0||nx>=n||ny<0||ny>=m) continue;
            if (maze[nx][ny]=='X' ) continue;
            if (maze[nx][ny]=='D'){
                if (time+1==t){
                    suc=true;
                    return;
                }
                else continue;
            }
            maze[nx][ny]='X';
            dfs(nx,ny,time+1);
            maze[nx][ny]='.';
            if (suc) return;
        }
        return;
    }
    
    int main (){
        
        while (cin>>n>>m>>t && !(m==0&&n==0&&t==0)){
            int sx,sy,dx,dy;
            suc=false;
            for (int i=0;i<n;i++){ 
                for (int j=0;j<m;j++){
                    cin>>maze[i][j];
                    if (maze[i][j]=='S'){
                        sx=i;sy=j;
                    }
                    if (maze[i][j]=='D'){
                        dx=i;dy=j;
                    }
                    
                }
            }
            if ((sx+sy)%2==((dx+dy)%2+t%2)%2){
                maze[sx][sy]='X';
                dfs(sx,sy,0);
            }
            string s=suc==true?"YES":"NO";
            cout<<s<<endl;
        }
    
        return 0;
    }

    提示让用scanf输入,其实不用也可以

  • 相关阅读:
    慎用SELECT INTO复制表
    Log4net 配置使用总结(一)
    清除Chrome浏览器的历史记录、缓存
    System.Runtime.InteropServices.COMException (0x80040154)错误
    MS SQL开发命名规则
    查看数据库、表、索引的物理存储情况
    (转)ASP.NET调用javascript脚本的方法总结
    SQL Server ——动态SQL
    SQL性能调优实践——SELECT COUNT
    养成随时注释的好习惯
  • 原文地址:https://www.cnblogs.com/yexiaoqi/p/7241699.html
Copyright © 2011-2022 走看看