zoukankan      html  css  js  c++  java
  • SCU4445——模拟——Right turn

    Right turn

    frog is trapped in a maze. The maze is infinitely large and divided into grids. It also consists of n obstacles, where the i-th obstacle lies in grid (xi,yi).

    frog is initially in grid (0,0), heading grid (1,0). She moves according to The Law of Right Turn: she keeps moving forward, and turns right encountering a obstacle.

    The maze is so large that frog has no chance to escape. Help her find out the number of turns she will make.

    Input

    The input consists of multiple tests. For each test:

    The first line contains 1 integer n (0n103). Each of the following n lines contains 2 integers xi,yi. (|xi|,|yi|109,(xi,yi)(0,0), all (xi,yi) are distinct)

    Output

    For each test, write 1 integer which denotes the number of turns, or ``-1'' if she makes infinite turns.

    Sample Input

     2
        1 0
        0 -1
        1
        0 1
        4
        1 0
        0 1
        0 -1
        -1 0

    Sample Output

     2
        0
        -1

    /*
       模拟题 机器人碰了就往右拐
       每次判断要与所有点判断并且是靠自己最近的点
    */
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    using namespace std;
    
    const int inf = 0x3f3f3f3f;
    int dx[] = { 1, 0, -1, 0};
    int dy[] = { 0, -1, 0, 1};
    int vis[1100][4];
    int obx[1100], oby[1100];
    int n;
    
    void work()
    {
        memset(vis, 0, sizeof(vis));
        int dir = 0;
        int nx , ny, dis;
        nx = ny = 0;
        for(int step = 0; ; step++){
            int mx = inf;
            dis = inf;
            int index = -1;
            for(int i = 1; i <= n ; i++){
                if(dir == 0 ){
                    if(ny == oby[i] && nx < obx[i]){
                        dis = obx[i] - nx - 1;
                    }
                }        
                else if(dir == 1){
                    if(nx == obx[i] && ny > oby[i]){
                    //   printf("%d ",i);
                        dis = ny - oby[i] - 1;
                    }
                }
                else if(dir == 2){
                    if(ny == oby[i] && nx > obx[i]){
                        dis = nx - obx[i] - 1;
                    }
                }
                else {
                    if(nx == obx[i] && ny < oby[i]){
                        dis = oby[i] - ny - 1;
                    }
                }
                if(dis < mx){
                    mx = dis;
                    index = i;
                }
            }
           // printf("%d %d %d
    ",index, step,dir);
        
          if(index == -1) {
                printf("%d
    ",step);
                return ;
            }
          if(vis[index][dir]){
              printf("-1
    ");
              return ;
          }
    
            vis[index][dir] = 1;
                if(dir == 0){
                    nx = obx[index] - 1; ny = oby[index];
                }
                else if(dir == 1){
                    nx = obx[index]; ny = oby[index] + 1;
                }
                else if(dir == 2){
                    nx = obx[index] + 1; ny = oby[index];
                }
                else {
                    nx = obx[index]; ny = oby[index] - 1;
                }
                //  printf("%d %d %d 
    ",nx,ny, dir);
            dir = (dir + 1) % 4;
        }
    }
    
    int main()
    {
        while(~scanf("%d", &n)){
            for(int i = 1; i <= n; i++){
                scanf("%d%d", &obx[i], &oby[i]);
            }
            work();
        }
        return 0;
    }
    

      

  • 相关阅读:
    将博客搬至CSDN
    java面试题(一)
    大数据面试题(二)
    大数据面试题(一)
    现在有T1、T2、T3三个线程,你怎样保证T2在T1执行完后执行,T3在T2执行完后执行?
    java序列化与反序列化总结
    Hashmap,Set,Map,List,ArrayList的区别
    Windows下安装Scrapy
    Eclipse java上手
    lucene文件检索(1)-linux java环境配置
  • 原文地址:https://www.cnblogs.com/zero-begin/p/4647005.html
Copyright © 2011-2022 走看看