POJ 1228 稳定凸包问题

来源：互联网发布：数据结构与算法的关系编辑：程序博客网时间：2024/06/16 03:06
题意：给出一些凸包上的点  这些点均是凸包上的点 判断该凸包是否为稳定凸包  即是否每条边上是否至少有三个点  改下凸包的模板即可#include <iostream>#include "stdio.h"#include "stdlib.h"#include "string.h"#include "math.h"#include "algorithm"#include <queue>using namespace std;#define modulo 1000000007const double eps=1e-8;int n,top;double ansx,ansy;struct ed {    double x,y;}point[1010],sta[1010];double mult(double x1,double y1,double x2,double y2,double x3,double y3){    return (x1-x3)*(y2-y3)-(y1-y3)*(x2-x3);}int  sgn(double x){    if(fabs(x)<eps) return 0;    if(x<0) return -1;    else return 1;}double dis(ed a,ed b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}int cmp(const void *a,const void *b){    ed c=*(ed *)a;    ed d=*(ed *)b;    double k=mult(c.x,c.y,d.x,d.y,point[0].x,point[0].y);    if(k<0||(!k&&dis(c,point[0])>dis(d,point[0]))) return 1;    else return -1;}void convex(){    for(int i=1;i<n;i++)    {        ed temp;        if(point[i].y<point[0].y||(point[i].y==point[0].y&&point[i].x<point[0].x)){            temp=point[i];            point[i]=point[0];            point[0]=temp;        }    }    qsort(point+1,n-1,sizeof(point[0]),cmp);    sta[0]=point[0];sta[1]=point[1];    top=1;    for(int i=2;i<n;i++)    {        while(top>=1&&mult(sta[top].x,sta[top].y,point[i].x,point[i].y,sta[top-1].x,sta[top-1].y)<0)  top--;        top++;        sta[top]=point[i];    }}bool judge(){    for(int i=1;i<top;i++)        if(mult(sta[i].x,sta[i].y,sta[i+1].x,sta[i+1].y,sta[i-1].x,sta[i-1].y)!=0&&mult(sta[i+1].x,sta[i+1].y,sta[i+2].x,sta[i+2].y,sta[i].x,sta[i].y)!=0)            return false;    return true;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=0;i<n;i++)            scanf("%lf%lf",&point[i].x,&point[i].y);        if(n<6) printf("NO\n");        else {            convex();            if(judge()) printf("YES\n");            else printf("NO\n");        }    }    return 0;}
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