poj 3335 3130 1279 判断是否有核

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Sample Input

24 0 0 0 1 1 1 1 08 0 0  0 2  1 2  1 1  2 1  2 2  3 2  3 0

Sample Output

YESNO

题意：

顺时针给出n个点，求这个多边形是否有核，有就输出yes，否则输出no

题解：

套模板

多边形核的理解：在此多边形里面放一个摄像头，它可以扫到多边形的任意一个点

做法：半平面切割法。依次选择直线，然后和多边形交，然后剔除在外面的点，最后得到一个核区域或者点（形象的理解就像在削苹果）

poj 3335

#include<math.h>#include<stdio.h>#include<algorithm>using namespace std;#define eps 1e-8const int MAXN=10017;int n;double r;int cCnt,curCnt;///最终切割得到的多边形的顶点数、暂存顶点个数struct point{    double x,y;};point points[MAXN],p[MAXN],q[MAXN];///初始多边形顶点（顺时针）、最终切割后多边形顶点、暂存顶点void getline(point x,point y,double &a,double &b,double &c) ///两点x、y确定一条直线a、b、c为其系数{    a=y.y-x.y;    b=x.x-y.x;    c=y.x*x.y-x.x*y.y;}void initial(){    for(int i=1;i<=n;i++)        p[i]=points[i];    p[n+1]=p[1];    p[0]=p[n];    cCnt=n;}point intersect(point x,point y,double a,double b,double c)///点x、y所在直线与ax+by+c=0的交点{    double u=fabs(a*x.x+b*x.y+c);    double v=fabs(a*y.x+b*y.y+c);    point pt;    pt.x=(x.x*v+y.x*u)/(u+v);    pt.y=(x.y*v+y.y*u)/(u+v);    return pt;}void cut(double a,double b ,double c){    curCnt=0;    for(int i=1;i<=cCnt;i++)    {        if(a*p[i].x+b*p[i].y+c>=0)///点代入线都大于0，说明此点都在这条直线某一边，不用切            q[++curCnt]=p[i];        else        {            if(a*p[i-1].x+b*p[i-1].y+c>0)///如果p[i-1]在直线的右侧的话                q[++curCnt]=intersect(p[i],p[i-1],a,b,c);            if(a*p[i+1].x+b*p[i+1].y+c>0)                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]=q[1];    p[0]=p[curCnt];    cCnt=curCnt;}void solve(){    ///注意：默认点是顺时针    initial();    double a,b,c;    for(int i=1;i<=n;i++){        getline(points[i],points[i+1],a,b,c);        cut(a,b,c);    }}int main(){    int T;    //freopen("in.txt","r",stdin);    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        for(int i=1;i<=n;i++)            scanf("%lf%lf",&points[i].x,&points[i].y);        points[n+1]=points[1];        solve();        puts(cCnt<1?"NO":"YES");    }    return 0;}

poj 3130

#include<math.h>#include<stdio.h>#include<algorithm>using namespace std;#define eps 1e-8const int MAXN=10017;int n;double r;int cCnt,curCnt;///最终切割得到的多边形的顶点数、暂存顶点个数struct point{    double x,y;};point points[MAXN],p[MAXN],q[MAXN];///初始多边形顶点（顺时针）、最终切割后多边形顶点、暂存顶点void getline(point x,point y,double &a,double &b,double &c) ///两点x、y确定一条直线a、b、c为其系数{    a=y.y-x.y;    b=x.x-y.x;    c=y.x*x.y-x.x*y.y;}void initial(){    for(int i=1;i<=n;i++)        p[i]=points[i];    p[n+1]=p[1];    p[0]=p[n];    cCnt=n;}point intersect(point x,point y,double a,double b,double c)///点x、y所在直线与ax+by+c=0的交点{    double u=fabs(a*x.x+b*x.y+c);    double v=fabs(a*y.x+b*y.y+c);    point pt;    pt.x=(x.x*v+y.x*u)/(u+v);    pt.y=(x.y*v+y.y*u)/(u+v);    return pt;}void cut(double a,double b ,double c){    curCnt=0;    for(int i=1;i<=cCnt;i++)    {        if(a*p[i].x+b*p[i].y+c>=0)///点代入线都大于0，说明此点都在这条直线某一边，不用切            q[++curCnt]=p[i];        else        {            if(a*p[i-1].x+b*p[i-1].y+c>0)///如果p[i-1]在直线的右侧的话                q[++curCnt]=intersect(p[i],p[i-1],a,b,c);            if(a*p[i+1].x+b*p[i+1].y+c>0)                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]=q[1];    p[0]=p[curCnt];    cCnt=curCnt;}void solve(){    ///注意：默认点是顺时针    initial();    double a,b,c;    for(int i=1;i<=n;i++){        getline(points[i],points[i+1],a,b,c);        cut(a,b,c);    }}void GuiZhengHua(){     ///规整化方向，逆时针变顺时针，顺时针变逆时针    for(int i=1;i<=n;i++) q[i]=points[n-i+1];    for(int i=1;i<=n;i++) points[i]=q[i];}int main(){    //freopen("in.txt","r",stdin);    while(scanf("%d",&n)&&n)    {        for(int i=1;i<=n;i++)            scanf("%lf%lf",&points[i].x,&points[i].y);        GuiZhengHua();        points[n+1]=points[1];        solve();        puts(cCnt<1?"0":"1");    }    return 0;}

poj 1279 计算内核面积

#include<math.h>#include<stdio.h>#include<algorithm>using namespace std;#define eps 1e-8const int MAXN=2000;int n;int cCnt,curCnt;///最终切割得到的多边形的顶点数、暂存顶点个数struct point{    double x,y;};point points[MAXN],p[MAXN],q[MAXN];///初始多边形顶点（顺时针）、最终切割后多边形顶点、暂存顶点void getline(point x,point y,double &a,double &b,double &c) ///两点x、y确定一条直线a、b、c为其系数{    a=y.y-x.y;    b=x.x-y.x;    c=y.x*x.y-x.x*y.y;}void initial(){    for(int i=1;i<=n;i++)        p[i]=points[i];    p[n+1]=p[1];    p[0]=p[n];    cCnt=n;}point intersect(point x,point y,double a,double b,double c)///点x、y所在直线与ax+by+c=0的交点{    double u=fabs(a*x.x+b*x.y+c);    double v=fabs(a*y.x+b*y.y+c);    point pt;    pt.x=(x.x*v+y.x*u)/(u+v);    pt.y=(x.y*v+y.y*u)/(u+v);    return pt;}void cut(double a,double b ,double c){    curCnt=0;    for(int i=1;i<=cCnt;i++)    {        if(a*p[i].x+b*p[i].y+c>=0)///点代入线都大于0，说明此点都在这条直线某一边，不用切            q[++curCnt]=p[i];        else        {            if(a*p[i-1].x+b*p[i-1].y+c>0)///如果p[i-1]在直线的右侧的话                q[++curCnt]=intersect(p[i],p[i-1],a,b,c);            if(a*p[i+1].x+b*p[i+1].y+c>0)                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]=q[1];    p[0]=p[curCnt];    cCnt=curCnt;}double solve(){    ///注意：默认点是顺时针    initial();    double a,b,c;    for(int i=1;i<=n;i++){        getline(points[i],points[i+1],a,b,c);        cut(a,b,c);    }    double area=0;    for(int i=1;i<=curCnt;i++)        area+=p[i].x*p[i+1].y-p[i+1].x*p[i].y;    return fabs(area/2.0);}int main(){    int T;    //freopen("in.txt","r",stdin);    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        for(int i=1;i<=n;i++)            scanf("%lf%lf",&points[i].x,&points[i].y);        points[n+1]=points[1];        printf("%.2lf\n",solve());    }    return 0;}

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