HDU 3685 Rotational Painting(凸包+重心)

果7發表於2013-11-09
AI

Rotational Painting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2378    Accepted Submission(s): 672


Problem Description
Josh Lyman is a gifted painter. One of his great works is a glass painting. He creates some well-designed lines on one side of a thick and polygonal glass, and renders it by some special dyes. The most fantastic thing is that it can generate different meaningful paintings by rotating the glass. This method of design is called “Rotational Painting (RP)” which is created by Josh himself. 

You are a fan of Josh and you bought this glass at the astronomical sum of money. Since the glass is thick enough to put erectly on the table, you want to know in total how many ways you can put it so that you can enjoy as many as possible different paintings hiding on the glass. We assume that material of the glass is uniformly distributed. If you can put it erectly and stably in any ways on the table, you can enjoy it. 

More specifically, if the polygonal glass is like the polygon in Figure 1, you have just two ways to put it on the table, since all the other ways are not stable. However, the glass like the polygon in Figure 2 has three ways to be appreciated. 

Pay attention to the cases in Figure 3. We consider that those glasses are not stable.
 

Input
The input file contains several test cases. The first line of the file contains an integer T representing the number of test cases. 

For each test case, the first line is an integer n representing the number of lines of the polygon. (3<=n<=50000). Then n lines follow. The ith line contains two real number xi and yi representing a point of the polygon. (xi, yi) to (xi+1, yi+1) represents a edge of the polygon (1<=i<n), and (xn,yn) to (x1, y1) also represents a edge of the polygon. The input data insures that the polygon is not self-crossed.
 

Output
For each test case, output a single integer number in a line representing the number of ways to put the polygonal glass stably on the table.
 

Sample Input
2
4
0 0
100 0
99 1
1 1
6
0 0
0 10
1 10
1 1
10 1
10 0





 



Sample Output




2
3

Hint
The sample test cases can be demonstrated by Figure 1 and Figure 2 in Description part. 
 





 



Source



2010 Asia Hangzhou Regional Contest



 








題目所求，給一個多邊形，問有多少種可以將多邊形放穩的方法。先求出重心，看重心對映到凸包的每兩條邊上，線上段上說明可以放。詳見程式碼。





題目地址：Rotational Painting








 AC程式碼：




#include<iostream>
#include<string>
#include<cstring>
#include<vector>
#include<iomanip>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<deque>
#include<cstdio>
using namespace std;
double eps=1e-8;

const int N=50005;
struct point
{
    double x,y;
} p[N],res[N],q[N];

bool mult(point sp,point ep,point op)
{
    return (sp.x-op.x)*(ep.y-op.y)>=(ep.x-op.x)*(sp.y-op.y);
}   //返回凸包上頂點的個數，不包括在凸包上的點

bool operator < (const point &l,const point &r)
{
    return l.y<r.y||(l.y==r.y&&l.x<r.x);
}

int tubao(point pnt[], int n)
{
    int i, len, top = 1;
    sort(pnt, pnt + n);
    if(n == 0) return 0;
    res[0] = pnt[0];
    if(n == 1) return 1;
    res[1] = pnt[1];
    if(n == 2) return 2;
    res[2] = pnt[2];
    for(i = 2; i < n; i ++)
    {
        while(top && mult(pnt[i], res[top], res[top-1])) top --;
        res[++top] = pnt[i];
    }
    len = top;
    res[++top] = pnt[n - 2];

    for(i = n - 3; i >= 0; i --)
    {
        while(top != len && mult(pnt[i], res[top], res[top-1]))
            top --;
        res[++top] = pnt[i];
    }
    return top;
}

bool dianmul(point a,point b,point c)  //c向ab線段作垂足落在ab上
{
    double x1,y1,x2,y2,x3,y3,x4,y4;
    x1=b.x-a.x,y1=b.y-a.y;
    x2=c.x-a.x,y2=c.y-a.y;
    x3=a.x-b.x,y3=a.y-b.y;
    x4=c.x-b.x,y4=c.y-b.y;
    if(x1*x2+y1*y2>eps&&x3*x4+y3*y4>eps) return true;
    return false;
}

point gravi(point p[],int n)   //求重心
{
    int i;
    double A=0, a;
    point t;
    t.x = t.y = 0;
    p[n] = p[0];
    for (i=0; i < n; i++)
    {
        a = p[i].x*p[i+1].y - p[i+1].x*p[i].y;
        t.x += (p[i].x + p[i+1].x) * a;
        t.y += (p[i].y + p[i+1].y) * a;
        A += a;
    }
    t.x /= A*3;
    t.y /= A*3;
    return t;
}

int main()
{
    int i,n,tes;
    point zhong;  //重心的座標
    cin>>tes;

    while(tes--)
    {
        cin>>n;
        for(i=0; i<n; i++)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        zhong=gravi(p,n);

        int len=tubao(p,n);
        res[len]=res[0];

        int cnt=0;
        for(i=0; i<len; i++)
        {
            if(dianmul(res[i],res[i+1],zhong))
            {
                cnt++;
            }
        }
        cout<<cnt<<endl;
    }
    return 0;
}

//203MS








            




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