zoukankan      html  css  js  c++  java
  • HDU 4452 模拟

    Running Rabbits

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 1787    Accepted Submission(s): 1252



    Problem Description
    Rabbit Tom and rabbit Jerry are running in a field. The field is an N×N grid. Tom starts from the up-left cell and Jerry starts from the down-right cell. The coordinate of the up-left cell is (1,1) and the coordinate of the down-right cell is (N,N)。A 4×4 field and some coordinates of its cells are shown below:

    The rabbits can run in four directions (north, south, west and east) and they run at certain speed measured by cells per hour. The rabbits can't get outside of the field. If a rabbit can't run ahead any more, it will turn around and keep running. For example, in a 5×5 grid, if a rabbit is heading west with a speed of 3 cells per hour, and it is in the (3, 2) cell now, then one hour later it will get to cell (3,3) and keep heading east. For example again, if a rabbit is in the (1,3) cell and it is heading north by speed 2,then a hour latter it will get to (3,3). The rabbits start running at 0 o'clock. If two rabbits meet in the same cell at k o'clock sharp( k can be any positive integer ), Tom will change his direction into Jerry's direction, and Jerry also will change his direction into Tom's original direction. This direction changing is before the judging of whether they should turn around.
    The rabbits will turn left every certain hours. For example, if Tom turns left every 2 hours, then he will turn left at 2 o'clock , 4 o'clock, 6 o'clock..etc. But if a rabbit is just about to turn left when two rabbit meet, he will forget to turn this time. Given the initial speed and directions of the two rabbits, you should figure out where are they after some time.
     

    Input
    There are several test cases.
    For each test case:
    The first line is an integer N, meaning that the field is an N×N grid( 2≤N≤20).
    The second line describes the situation of Tom. It is in format "c s t"。c is a letter indicating the initial running direction of Tom, and it can be 'W','E','N' or 'S' standing for west, east, north or south. s is Tom's speed( 1≤s<N). t means that Tom should turn left every t hours( 1≤ t ≤1000).
    The third line is about Jerry and it's in the same format as the second line.
    The last line is an integer K meaning that you should calculate the position of Tom and Jerry at K o'clock( 1 ≤ K ≤ 200).
    The input ends with N = 0.
     

    Output
    For each test case, print Tom's position at K o'clock in a line, and then print Jerry's position in another line. The position is described by cell coordinate.
     

    Sample Input
    4 E 1 1 W 1 1 2 4 E 1 1 W 2 1 5 4 E 2 2 W 3 1 5 0
     

    Sample Output
    2 2 3 3 2 1 2 4 3 1 4 1
     

    Source
    2012 Asia JinHua Regional Contest

    一个兔子在左上角,另一个兔子在右下角,给定N*N矩阵,初始方向,速度,左转周期,和过了m秒,求m秒后两个兔子的坐标。注意当兔子相遇后交换方向,此时不左转。相遇是以每小时为单位后查看是否相遇,中间(在一小时内)相遇不算不反转。

    看到这个题我想到了以前做的一个搜索题,那个转向也是有头方向和四面方向的,当时学姐讲的我都没听懂,看到老胡代码豁然开朗,就是按照顺时针方向定义0,1,2,3,每次后转就直接+2,向左转+1,在定义两个x,y的移动数组对应固定北西南东四个方向,方向数组和移动数组一起用就解决这个问题了。

    老胡传送门:http://www.cnblogs.com/pach/p/5779371.html

    #include <iostream>
    #include <cstdio>
    #include <algorithm>
    using namespace std;
    
    int dx[]= {0,-1,0,1}; //方向顺序为N,W,S,E。这是为了方便向左转
    int dy[]= {-1,0,1,0};
    int n;
    int time;
    
    struct Coo
    {
        int x;
        int y;
    } coo[2];
    struct Rabbit
    {
        int di;
        int v;
        int turn;
    } rabbit[2];
    
    void solve()
    {
        coo[0].x=coo[0].y=1;
        coo[1].x=coo[1].y=n;
        for(int i=1; i<=time; i++)
        {
            bool flag1=true;
            for(int j=0; j<rabbit[0].v; j++) //一步一步走,不能往前走转向
            {
                int x=coo[0].x;
                int y=coo[0].y;
                int di=rabbit[0].di;//简化代码,方便书写(听老胡是这样,代码本身没什么意义,就是偷点懒)
                
                if((x==1 && di==1) || (x==n && di==3) || (y==1 && di==0) || (y==n && di==2))
                    rabbit[0].di=di=(di+2)%4; //向后转,方向逆时针转向,所以直接+2
                coo[0].x=x+dx[di];
                coo[0].y=y+dy[di];
            }
            for(int j=0; j<rabbit[1].v; j++)
            {
                int x=coo[1].x;
                int y=coo[1].y;
                int di=rabbit[1].di;
                if((x==1 && di==1) || (x==n && di==3) || (y==1 && di==0) || (y==n && di==2))
                    rabbit[1].di=di=(di+2)%4;
                coo[1].x=x+dx[di];
                coo[1].y=y+dy[di];
            }
            if(coo[0].x==coo[1].x && coo[0].y==coo[1].y)
            {
                int direct;
                direct=rabbit[0].di;
                rabbit[0].di=rabbit[1].di;
                rabbit[1].di=direct;
                flag1=false;
            }
            if(i%rabbit[0].turn==0 && flag1)
                rabbit[0].di=(rabbit[0].di+1)%4; //向左转
            if(i%rabbit[1].turn==0 && flag1)
                rabbit[1].di=(rabbit[1].di+1)%4;
        }
        printf("%d %d
    %d %d
    ",coo[0].y,coo[0].x,coo[1].y,coo[1].x);
    }
    int main()
    {
        //freopen("in.txt","r",stdin);
        while(scanf("%d",&n),n)
        {
            for(int i=0; i<2; i++)
            {
                char c;
                getchar();
                scanf("%c%d%d",&c,&rabbit[i].v,&rabbit[i].turn);
                if(c=='N')rabbit[i].di=0;       //将方向转化为数字,向左转的话+1再模4,向后转+2再模4
                else if(c=='W')rabbit[i].di=1;
                else if(c=='S')rabbit[i].di=2;
                else if(c=='E')rabbit[i].di=3;
            }
            scanf("%d",&time);
            solve();
        }
        return 0;
    }



  • 相关阅读:
    淘宝npm镜像
    mousedown监听onmousemove判断鼠标指针移动方向的JavaScript
    vue 生命周期流程图 go.js
    微信在pc端打开多窗口.bat
    bootstrap配置
    前端入门到进阶图文教程,超详细的Web前端学习笔记。从零开始学前端,做一名精致优雅的前端工程师。
    低代码开发平台
    开源在线绘图工具,界面美观,功能丰富
    IOS 空字符串报错 解决办法
    vue 在使用数组的时候,数组内部数据发生变化,视图却没有改变
  • 原文地址:https://www.cnblogs.com/zhangmingzhao/p/7256737.html
Copyright © 2011-2022 走看看