几何专项2D：POJ 2318

许久不做几何，完全不会了。刚开始O(n^2)纯暴力，妥妥地T了……其实只要利用叉积的性质，判断该点在隔板的左边还是右边，然后二分答案即可。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn=5010;struct point{    int x,y;    point(int x=0,int y=0):x(x),y(y){}};point operator-(point a,point b){return point(a.x-b.x,a.y-b.y);}point operator+(point a,point b){return point(a.x+b.x,a.y+b.y);}point operator*(point a,double p){return point(a.x*p,a.y*p);}int cross(point a,point b){return a.x*b.y-a.y*b.x;}int dot(point a,point b){return a.x*b.x+a.y*b.y;}bool onseg(point p,point a1,point a2){    return cross(a1-p,a2-p)==0 && dot(a1-p,a2-p)<0;}int sign(int x){    if(x==0) return 0;    else return x<0?-1:1;}point poly[maxn][4];int isinpoly(point p,int u){    int wn=0;    for(int i=0;i<4;i++)    {        if(onseg(p,poly[u][i],poly[u][(i+1)%4])) return 1;        int k=sign(cross(poly[u][(i+1)%4]-poly[u][i],p-poly[u][i]));        int d1=sign(poly[u][i].y-p.y);        int d2=sign(poly[u][(i+1)%4].y-p.y);        if(k>0 && d1<=0 && d2>0) wn++;        if(k<0 && d2<=0 && d1>0) wn--;    }    if(wn!=0) return 1;    return 0;}int cnt[maxn];int n,m,x0,y0,x1,y1;int main(){    //freopen("in.txt","r",stdin);    int flag=0;    while(scanf("%d",&n)&&n)    {        scanf("%d%d%d%d%d",&m,&x0,&y0,&x1,&y1);        if(flag) puts("");        else flag=1;        memset(cnt,0,sizeof(cnt));        point p1=point(x0,y0),p2=point(x0,y1);        for(int i=0;i<n;i++)        {            point t1,t2;            scanf("%d%d",&t1.x,&t2.x);            t1.y=y0;t2.y=y1;            poly[i][0]=p1;poly[i][1]=t1;poly[i][2]=t2;poly[i][3]=p2;            p1=t1;p2=t2;        }        poly[n][0]=p1;poly[n][1]=point(x1,y0);poly[n][2]=point(x1,y1);poly[n][3]=p2;        while(m--)        {            point p;            scanf("%d%d",&p.x,&p.y);            //cout<<p.x<<" "<<p.y<<endl;            int L=0,R=n;            while(L<R)            {                int mid=L+(R-L+1)/2;                if(cross(poly[mid][0]-poly[mid][3],p-poly[mid][3])>0) R=mid-1;                else L=mid;            }            cnt[L]++;        }        for(int i=0;i<=n;i++)            printf("%d: %d\n",i,cnt[i]);    }    return 0;}