zoukankan      html  css  js  c++  java
  • Swordsman ACM HDU 3902(判断n边形是不是轴对称图形,暴力求解了~~~~)

    Swordsman

    Time Limit: 10000/4000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)
    Total Submission(s): 308    Accepted Submission(s): 116


    Problem Description
    Mr. AC is a swordsman. His dream is to be the best swordsman in the world. To achieve his goal, he practices every day. There are many ways to practice, but Mr. AC likes “Perfect Cut” very much. The “Perfect Cut” can be described as the following two steps:
    1.  Put a piece of wood block on the desk, and then suddenly wave the sword, cutting the block into two pieces.
    2.  Without any motion, the two pieces must be absolutely axial symmetry.
    According to the step two, when the board is an axial symmetry figure, Mr. AC has a chance to achieve the “Perfect Cut”. Now give you a board, and you should tell if Mr. AC has a chance to complete the “Perfect Cut”. The board is a simple polygon.
     

    Input
    The input contains several cases.
    The first line of one case contains a single integer n (3 <= n <= 20000), the number of points. The next n lines indicate the points of the simple polygon, each line with two integers x, y (0 <= x, y <= 20000). The points would be given either clockwise or counterclockwise.
     

    Output
    For each case, output the answer in one line. If Mr. AC has the chance, print “YES”, otherwise "NO".
     

    Sample Input
    3 0 0 2 0 0 1 4 0 0 0 1 1 1 1 0
     

    Sample Output
    NO YES
    Hint
    Huge input, scanf is recommended.
     

    Source
     

    Recommend
    xubiao
     
     
    暴力求解,还没有发现好的方法。
    /*
    判断n边形是否是轴对称图形:
    n边形的n个顶点再加上n条边的中点,共2*n个顶点。
    如果存在对称轴,必定是点i和点i+n连成的直线,然后分别验证两边对称的点
    到点i和点i+n的距离是否相等
    */

    #include
    <stdio.h>
    #include
    <cmath>
    using namespace std;
    #define MAXN 20000
    #define eps 1e-5
    struct Node
    {
    double x,y;
    }node[
    2*MAXN+10];
    int n,m;
    bool flag;//用来标注是否是轴对称图形
    double dis(int i,int j)//求node[i]和node[j]的距离
    {
    double x=node[i].x-node[j].x;
    double y=node[i].y-node[j].y;
    return sqrt(x*x+y*y);
    }
    bool check(int i,int j,int x,int y)//检查node[i]和node[j]是否关于xy对称
    //对称则dis(i,x)==dis(j,x)&&dis(i,y)==dis(j,y);
    {
    if(fabs(dis(i,x)-dis(j,x))>eps) return false;
    if(fabs(dis(i,y)-dis(j,y))>eps) return false;
    return true;
    }
    void ff(int x,int y)//判断node[x]和node[y]组成的直线是不是对称轴
    {
    int i,j;
    i
    =j=x;
    while(1)
    {
    i
    ++;j--;
    if(j==0) j=m;
    if(i==y)
    {
    flag
    =true;
    return;
    }
    if(check(i,j,x,y)==false) return;
    }
    }
    int main()
    {
    //freopen("test.in","r",stdin);
    //freopen("test.out","w",stdout);
    int i;
    while(~scanf("%d",&n))
    {
    m
    =2*n;
    for(i=1;i<=m;i+=2)
    {
    scanf(
    "%lf%lf",&node[i].x,&node[i].y);
    }
    node[m
    +1]=node[1];
    for(i=2;i<=m;i+=2)
    {
    node[i].x
    =(node[i-1].x+node[i+1].x)/2;
    node[i].y
    =(node[i-1].y+node[i+1].y)/2;
    }
    flag
    =false;
    for(i=1;i<=n;i++)
    {
    ff(i,i
    +n);
    if(flag) break;
    }
    if(flag) printf("YES\n");
    else printf("NO\n");
    }
    return 0;
    }

  • 相关阅读:
    学习素材、网站
    用 Python脚本生成 Android SALT 扰码
    H面试程序(29):求最大递增数
    常用数据库查询判断表和字段是否存在
    《火球——UML大战需求分析》(第3章 分析业务模型-类图)——3.7 关于对象图
    N个数依次入栈,出栈顺序有多少种
    WIN ERROR:C:WindowsSystem32<LANG_NAME>mstsc.exe.MUI
    大端法和小端法
    freopen()重定向的打开和关闭
    Linux 的 Spinlock 在 MIPS 多核处理器中的设计与实现
  • 原文地址:https://www.cnblogs.com/kuangbin/p/2126194.html
Copyright © 2011-2022 走看看