zoukankan      html  css  js  c++  java
  • upc组队赛5 Hunter’s Apprentice 【判断多边形边界曲线顺逆时针】

    Hunter’s Apprentice

    题目描述

    When you were five years old, you watched in horror as a spiked devil murdered your parents. You would have died too, except you were saved by Rose, a passing demon hunter. She ended up adopting you and training you as her apprentice.
    Rose’s current quarry is a clock devil which has been wreaking havoc on the otherwise quiet and unassuming town of Innsmouth. It comes out each night to damage goods, deface signs, and kill anyone foolish enough to wander around too late. The clock devil has offed the last demon hunter after it; due to its time-warping powers, it is incredibly agile and fares well in straight-up fights.
    The two of you have spent weeks searching through dusty tomes for a way to defeat this evil. Eventually, you stumbled upon a relevant passage. It detailed how a priest managed to ensnare a clock devil by constructing a trap from silver, lavender, pewter, and clockwork. The finished trap contained several pieces, which must be placed one-by-one in the shape of a particular polygon, in counter-clockwise order. The book stated that the counter-clockwise order was important to counter the clock devil’s ability to speed its own time up, and that a clockwise order would only serve to enhance its speed.
    It was your responsibility to build and deploy the trap, while Rose prepared for the ensuing fight. You carefully reconstruct each piece as well as you can from the book. Unfortunately, things did not go as planned that night. Before you can finish preparing the trap, the clock devil finds the two of you. Rose tries to fight the devil, but is losing quickly. However, she is buying you the time to finish the trap. You quickly walk around them in the shape of the polygon, placing each piece in the correct position. You hurriedly activate the trap as Rose is knocked out. Just then, you remember the book’s warning. What should you do next?

    输入

    The first line of input contains a single integer T (1 ≤ T ≤ 100), the number of test cases. The first line of each test case contains a single integer n (3 ≤ n ≤ 20), the number of pieces in the trap. Each of the next n lines contains two integers xi and yi (|xi |, |yi | ≤ 100), denoting the x and y coordinates of where the ith piece was placed. It is guaranteed that the polygon is simple; edges only intersect at vertices, exactly two edges meet at each vertex, and all vertices are distinct.

    输出

    For each query in the input, output a single line containing the number of Trapezoid troops dropped in that city.

    样例输入

    2
    3
    0 0
    1 0
    0 1
    3
    0 0
    0 1
    1 0
    

    样例输出

    fight
    run
    

    提示

    In the first case, you went around the polygon in the correct direction, so it is safe to fight the clock devil
    and try to save Rose.
    In the second case, you messed up, and it is time to start running. Sorry Rose!

    题意

    你要去帮罗斯去打怪兽balabala,然后你得绕着一个多边形跑,逆时针能帮上忙,顺时针会帮倒忙

    给你一堆点坐标,然后按点坐标顺序跑,判断是顺时针还是逆时针

    这有个公式,如果知道公式就是简单题了,

    如果不知道又像我这种一辈子都不可能手推出来公式的就只能表示 模板+1

    关于公式 https://blog.csdn.net/henuyh/article/details/80378818

    代码

    #include<bits/stdc++.h>
    using namespace std;
    #define pb push_back
    #define mp make_pair
    #define rep(i,a,n) for(int i=a;i<n;++i)
    #define readc(x) scanf("%c",&x)
    #define read(x) scanf("%d",&x)
    #define sca(x) scanf("%d",&x)
    #define read2(x,y) scanf("%d%d",&x,&y)
    #define read3(x,y,z) scanf("%d%d%d",&x,&y,&z)
    #define print(x) printf("%d
    ",x)
    #define mst(a,b) memset(a,b,sizeof(a))
    #define lowbit(x) x&-x
    #define lson(x) x<<1
    #define rson(x) x<<1|1
    #define pb push_back
    #define mp make_pair
    typedef pair<int,int> P;
    typedef long long ll;
    const int INF =0x3f3f3f3f;
    const int inf =0x3f3f3f3f;
    const int mod = 1e9+7;
    const int MAXN = 105;
    const int maxn = 10010;
    double x[1050],y[1050];
    int main()
    {
      int t;
      read(t);
      while(t--)
      {
        int n;
        read(n);
        double temp = 0;
        rep(i, 0, n)
        {
          scanf("%lf%lf", &x[i], &y[i]);
        }
        rep(i,0,n-1)
        {
          temp += -0.5 * (y[i + 1] + y[i]) * (x[i + 1] - x[i]); //Green 公式
        }
        if(temp > 0)
          cout<<"fight"<<endl;
        else
          cout<<"run"<<endl;
      }
      return 0;
    }
    
  • 相关阅读:
    【算法】HashMap相关要点记录
    【算法】二叉树、N叉树先序、中序、后序、BFS、DFS遍历的递归和迭代实现记录(Java版)
    SpringCloud Openfeign Get请求服务传递对象的报400 Post not support的错误解决办法
    掌握 Promise 的逻辑方法
    JavaScript的执行上下文,真没你想的那么难
    一套标准的ASP.NET Core容器化应用日志收集分析方案
    在IIS中部署前后端应用,多么痛的领悟!
    吐槽一下Abp的用户和租户管理模块
    ant-design-vue中tree增删改
    微服务下的注册中心如何选择
  • 原文地址:https://www.cnblogs.com/llke/p/10800004.html
Copyright © 2011-2022 走看看