POJ 1228 Grandpa's Estate(凸包)

Grandpa's Estate

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11289		Accepted: 3117

Description

Being the only living descendant of his grandfather, Kamran the Believer inherited all of the grandpa's belongings. The most valuable one was a piece of convex polygon shaped farm in the grandpa's birth village. The farm was originally separated from the neighboring farms by a thick rope hooked to some spikes (big nails) placed on the boundary of the polygon. But, when Kamran went to visit his farm, he noticed that the rope and some spikes are missing. Your task is to write a program to help Kamran decide whether the boundary of his farm can be exactly determined only by the remaining spikes.

Input

The first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains an integer n (1 <= n <= 1000) which is the number of remaining spikes. Next, there are n lines, one line per spike, each containing a pair of integers which are x and y coordinates of the spike.

Output

There should be one output line per test case containing YES or NO depending on whether the boundary of the farm can be uniquely determined from the input.

Sample Input

1
6 
0 0
1 2
3 4
2 0
2 4 
5 0

Sample Output

NO

Source

Tehran 2002 Preliminary

题目大意: 给定n个点, 都是凸包上的顶点, 问这个凸包是否为稳定凸包,就是唯一确定一个凸包

思路: 刚开始老想着怎么判定它是否是凸包就完事了, 交了好几次都是wa, 后来去网上一看, 原来不是这样的。。。是求稳定凸包的，如果给定的顶点唯一确定的一个凸包的话， 那么这个多边形的每一条边上的顶点个数一定是大于等于三个的(包括端点), 只有这样才能保证是稳定的。如图所示。

图一中， 如果AB之间没有第三个点的话， 那么就有可能出现一个点E还使得原图形为凸包， 所以它是不稳定的， 图二就是稳定的了，有了这些点，他可以唯一确定一个凸包。知道了这个之后写代码就容易多了。

我的代码：

/*************************************************************************
    > File Name:            poj_1228.cpp
    > Author:               Howe_Young
    > Mail:                 1013410795@qq.com
    > Created Time:         2015年05月07日 星期四 16时44分36秒
 ************************************************************************/

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#define EPS 1e-8
using namespace std;
const int maxn = 1004;
struct point{
    double x, y;
};
point p[maxn], convex[maxn];
double Min(double a, double b)
{
    return a < b ? a : b;
}
double Max(double a, double  b)
{
    return a > b ? a : b;
}
int sgn(double x)
{
    if (fabs(x) < EPS)
        return 0;
    return x < 0 ? -1 : 1;
}
double x_multi(point p1, point p2, point p3)
{
    return (p3.x - p1.x) * (p2.y - p1.y) - (p2.x - p1.x) * (p3.y - p1.y);
}
bool cmp(const point p1, const point p2)
{
    return ((p1.y == p2.y && p1.x < p2.x) || p1.y < p2.y);
}
void convex_hull(point *p, point *convex, int n, int &len)
{
    sort(p, p + n, cmp);
    int top = 1;
    convex[0] = p[0];
    convex[1] = p[1];
    //找出凸包的下半部分凸壳 
    for (int i = 2; i < n; i++)
    {
        while (top > 0 && sgn(x_multi(convex[top - 1], convex[top], p[i])) <= 0)//大于0为逆时针,小于0为顺时针
            top--;
        convex[++top] = p[i];
    }
    int tmp = top;
    //找出凸包的上半部分，因为我的比较函数是写的y优先的，所以上下部分，当然也可以以x优先排序，这时候就是左右部分了 
    for (int i = n - 2; i >= 0; i--)
    {
        while (top > tmp && sgn(x_multi(convex[top - 1], convex[top], p[i])) <= 0)//大于0为逆时针,小于0为顺时针
            top--;
        convex[++top] = p[i];//存放凸包中的点
    }
    len = top;
}
bool on_segment(point p1, point p2, point p3)//判断p3是否在线段p1p2上
{
    double minx, miny, maxx, maxy;
    minx = Min(p1.x, p2.x);
    maxx = Max(p1.x, p2.x);
    miny = Min(p1.y, p2.y);
    maxy = Max(p1.y, p2.y);
    return (sgn(x_multi(p1, p2, p3)) == 0 && (sgn(p3.x - minx) >= 0 && sgn(p3.x - maxx) <= 0 && sgn(p3.y - miny) >= 0 && sgn(p3.y - maxy) <= 0));
}
bool check(point *p, point p1, point p2, int n)
{
    int cnt = 0;
    for (int i = 0; i < n; i++)
    {
        if (on_segment(p1, p2, p[i]))
            cnt++;
    }
    if (cnt == n)//特判，如果给定的点成一条线时，不符合
        return false;
    return cnt >= 3;
}
int main()
{
    int kase, n;
    scanf("%d", &kase);
    while (kase--)
    {
        scanf("%d", &n);
        for (int i = 0; i < n; i++)
            scanf("%lf %lf", &p[i].x, &p[i].y);
        int len;
        if (n <= 5)//n <= 5之前的点都是不确定的
        {
            puts("NO");
            continue;
        }
        convex_hull(p, convex, n, len);
        convex[len] = convex[0];
        bool flag = false;
        for (int i = 0; i < len; i++)//检查凸包中的每一个边
        {
            if (!check(p, convex[i], convex[i + 1], n))
            {
                flag = true;
                break;
            }
        }
        if (!flag)
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}