hdu 5533 Dancing Stars on Me 水題

時間 2021-08-14

標籤 php ios app ide this spa orm three 欄目 PHP 简体版

原文原文鏈接

Dancing Stars on Me

Time Limit: 20 Secphp

Memory Limit: 256 MBios

題目鏈接

http://acm.hdu.edu.cn/showproblem.php?pid=5533app

Description

The sky was brushed clean by the wind and the stars were cold in a black sky. What a wonderful night. You observed that, sometimes the stars can form a regular polygon in the sky if we connect them properly. You want to record these moments by your smart camera. Of course, you cannot stay awake all night for capturing. So you decide to write a program running on the smart camera to check whether the stars can form a regular polygon and capture these moments automatically.

Formally, a regular polygon is a convex polygon whose angles are all equal and all its sides have the same length. The area of a regular polygon must be nonzero. We say the stars can form a regular polygon if they are exactly the vertices of some regular polygon. To simplify the problem, we project the sky to a two-dimensional plane here, and you just need to check whether the stars can form a regular polygon in this plane.

⋅1. If you touch a buoy before your opponent, you will get one point. For example if your opponent touch the buoy #2 before you after start, he will score one point. So when you touch the buoy #2, you won't get any point. Meanwhile, you cannot touch buoy #3 or any other buoys before touching the buoy #2.

⋅2. Ignoring the buoys and relying on dogfighting to get point. If you and your opponent meet in the same position, you can try to fight with your opponent to score one point. For the proposal of game balance, two players are not allowed to fight before buoy #2 is touched by anybody.

There are three types of players.

Speeder: As a player specializing in high speed movement, he/she tries to avoid dogfighting while attempting to gain points by touching buoys.
Fighter: As a player specializing in dogfighting, he/she always tries to fight with the opponent to score points. Since a fighter is slower than a speeder, it's difficult for him/her to score points by touching buoys when the opponent is a speeder.
All-Rounder: A balanced player between Fighter and Speeder.

There will be a training match between Asuka (All-Rounder) and Shion (Speeder).
Since the match is only a training match, the rules are simplified: the game will end after the buoy #1 is touched by anybody. Shion is a speed lover, and his strategy is very simple: touch buoy #2,#3,#4,#1 along the shortest path.

Asuka is good at dogfighting, so she will always score one point by dogfighting with Shion, and the opponent will be stunned for T seconds after dogfighting. Since Asuka is slower than Shion, she decides to fight with Shion for only one time during the match. It is also assumed that if Asuka and Shion touch the buoy in the same time, the point will be given to Asuka and Asuka could also fight with Shion at the buoy. We assume that in such scenario, the dogfighting must happen after the buoy is touched by Asuka or Shion.

The speed of Asuka is V1 m/s. The speed of Shion is V2 m/s. Is there any possibility for Asuka to win the match (to have higher score)?

Inputide

The first line contains a integer T indicating the total number of test cases. Each test case begins with an integer n, denoting the number of stars in the sky. Following n lines, each contains 2 integers xi,yi, describe the coordinates of n stars.

1≤T≤300
3≤n≤100
−10000≤xi,yi≤10000
All coordinates are distinct.

Outputthis

For each test case, please output "`YES`" if the stars can form a regular polygon. Otherwise, output "`NO`" (both without quotes).

Sample Inputspa

Sample Outputorm

NO
YES
NO

HINT

題意
three

給你n個整數點，而後問你是否這幾個點可以構成一個正多邊形ip

題解：ci

只用考慮n=4的狀況，而後判斷是否爲一個正方形就行了

而後瞎搞一波。。。

代碼

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<cstring>
using namespace std;

pair<int,int> p[302];
int check()
{
    sort(p,p+4);
    vector<int>G;
    for(int i=0;i<4;i++)
        for(int j=i+1;j<4;j++)
        {
            int x = p[i].first - p[j].first;
            int y = p[i].second - p[j].second;
            G.push_back((x*x)+(y*y));
        }
    sort(G.begin(),G.end());
    for(int i=1;i<4;i++)
        if(G[i]!=G[i-1])
            return 0;
    if(G[4]==G[3])return 0;
    if(G[5]!=G[4])return 0;
    return 1;
}
int main()
{
    int t;scanf("%d",&t);
    while(t--)
    {
        int n;scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d%d",&p[i].first,&p[i].second);
        if(n!=4)
        {
            printf("NO\n");continue;
        }
        sort(p,p+4);
        if(check())printf("YES\n");
        else printf("NO\n");
    }
}

Dancing Stars on Me

Time Limit: 20 Sec

Memory Limit: 256 MB

題目鏈接

http://acm.hdu.edu.cn/showproblem.php?pid=5533

Description

Input

Output

For each test case, please output "`YES`" if the stars can form a regular polygon. Otherwise, output "`NO`" (both without quotes).

Sample Input

Sample Output

NO
YES
NO

HINT

題意

給你n個整數點，而後問你是否這幾個點可以構成一個正多邊形

題解：

只用考慮n=4的狀況，而後判斷是否爲一個正方形就行了

而後瞎搞一波。。。

代碼

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<cstring>
using namespace std;

pair<int,int> p[302];
int check()
{
    sort(p,p+4);
    vector<int>G;
    for(int i=0;i<4;i++)
        for(int j=i+1;j<4;j++)
        {
            int x = p[i].first - p[j].first;
            int y = p[i].second - p[j].second;
            G.push_back((x*x)+(y*y));
        }
    sort(G.begin(),G.end());
    for(int i=1;i<4;i++)
        if(G[i]!=G[i-1])
            return 0;
    if(G[4]==G[3])return 0;
    if(G[5]!=G[4])return 0;
    return 1;
}
int main()
{
    int t;scanf("%d",&t);
    while(t--)
    {
        int n;scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d%d",&p[i].first,&p[i].second);
        if(n!=4)
        {
            printf("NO\n");continue;
        }
        sort(p,p+4);
        if(check())printf("YES\n");
        else printf("NO\n");
    }
}