POJ_3335_Rotating Scoreboard(半平面交的核)

来源：互联网发布：朱棣是朱元璋亲生知乎编辑：程序博客网时间：2024/06/10 04:56

题型：计算几何

题意：判断多边形是否存在核

分析：半平面交模板题

代码：

#include<iostream>#include<cstring>#include<cmath>#include<cstdio>#define mt(a,b) memset(a,b,sizeof(a))using namespace std;const int maxn = 1234;const int inf=0x7ffffff;const double PI=acos(-1.0);const double eps=1e-8;const double e=2.7182818284590452354;struct point {    double x,y;    point() {}    point(double x,double y): x(x),y(y) {}} p[maxn],q[maxn],pnt[maxn];int cnt,curcnt;void initial(int n) {    for(int i=1; i<=n; i++)        p[i]=pnt[i];    p[n+1]=p[1];    p[0]=p[n];    cnt=n;}//两点确定直线void getline(point x,point y,double &a,double &b,double &c) {    a=y.y-x.y;    b=x.x-y.x;    c=y.x*x.y-x.x*y.y;}//求交点point intersect(point x,point y,double a,double b,double c) {    double u=fabs(a*x.x+b*x.y+c);    double v=fabs(a*y.x+b*y.y+c);    return point((v*x.x+u*y.x)/(u+v),(v*x.y+u*y.y)/(u+v));}//直线切割void cut(double a,double b,double c) {    curcnt=0;    for(int i=1; i<=cnt; i++) {        if(a*p[i].x+b*p[i].y+c>=0)//当前点在直线右侧            q[++curcnt]=p[i];        else {            if(a*p[i-1].x+b*p[i-1].y+c>0)//前一个点在直线右侧                q[++curcnt]=intersect(p[i],p[i-1],a,b,c);            if(a*p[i+1].x+b*p[i+1].y+c>0)//同理                q[++curcnt]=intersect(p[i],p[i+1],a,b,c);        }    }    for(int i=1; i<=curcnt; i++)        p[i]=q[i];    p[curcnt+1]=p[1];    p[0]=p[curcnt];    cnt=curcnt;}int main() {    int n,t;    scanf("%d",&t);    while(t--) {        scanf("%d",&n);        for(int i=1; i<=n; i++)            scanf("%lf%lf",&pnt[i].x,&pnt[i].y);        pnt[n+1]=pnt[1];        initial(n);        //顺时针方向        for(int i=1; i<=n; i++) {            double a,b,c;            getline(pnt[i],pnt[i+1],a,b,c);            cut(a,b,c);        }        if(cnt<1) puts("NO");        else puts("YES");    }    return 0;}

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