POJ - 1228 Grandpa's Estate (稳定凸包 )

最新推荐文章于 2021-07-26 21:44:39 发布

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本文介绍了一种算法，用于帮助确定缺失部分边界标记后，农场边界是否仍可唯一确定。通过输入剩余的钉子坐标，算法首先计算凸包，然后判断凸包是否为稳定凸包，即所有边界线上点的数量是否大于2。

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Grandpa's Estate

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

Sample Output

NO

题目链接：
题目大意：给n个点，求凸包，然后判断此凸包是否是稳定凸包

思路：稳定凸包：凸包的所有边上点的个数都大于2，先求凸包，后判断凸包上是否存在一条边只有两个端点

代码：

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
using namespace std;
#define ll long long
const double pi=acos(-1.0);
const int N=10010;
struct node
{
    ll x,y;
} e[N],s[N],a[N];
bool cmp(node a,node b)
{
    if(a.y==b.y) return a.x<b.x;
    return a.y<b.y;
}
ll cross(node a,node b,node c)
{
    return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
}
ll dot(node a,node b,node c)
{
    return (b.x-a.x)*(c.x-a.x)+(b.y-a.y)*(c.y-a.y);
}
double dis(node a,node b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
bool cmp1(node a,node b)
{
    ll mm=cross(e[0],a,b);
    if(mm>0) return 1;
    else if(mm==0&&dis(e[0],a)-dis(e[0],b)<=0)
        return 1;
    else return 0;
}
int n,tot;
void graham()
{
    sort(e,e+n,cmp);
    a[0]=e[0];
    int l=1;
    for(int i=1; i<n; i++)
    {
        if(e[i].x==e[i-1].x&&e[i].y==e[i-1].y) continue;
        a[l++]=e[i];
    }
    n=l;
    if(n<3){tot=n-1;return;}
    sort(a+1,a+n,cmp1);
    s[0]=a[0],s[1]=a[1];
    tot=1;
    for(int i=2; i<n; i++)
    {
        while(tot&&cross(s[tot-1],s[tot],a[i])<0)
            tot--;
        s[++tot]=a[i];
    }
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0; i<n; i++)
            scanf("%lld%lld",&e[i].x,&e[i].y);
        if(n<6) printf("NO\n");
        else
        {
            graham();
            int flag=0;
            for(int i=1; i<tot; i++)
            {
                if(cross(s[i-1],s[i],s[i+1])!=0&&cross(s[i],s[i+1],s[i+2])!=0)
                {
                    flag=1;
                    break;
                }
            }
            if(flag) printf("NO\n");
            else printf("YES\n");
        }
    }
    return 0;
}