POJ2318 TOYS 和POJ2398 Toy Storage题解（点在四边形内）（简单几何）

POJ 2318 TOYS 链接：http://poj.org/problem?id=2318

POJ 2398 Toy Storage 链接：http://poj.org/problem?id=2398

poj2318题目大意：给你一个盒子的俯视图，从左到右将每个格子划分为0,1,2...n;给你一些点的坐标，让你输出每个格子里点的个数。

poj2398题目大意：给你一个盒子的俯视图，从左到右将每个格子划分为0,1,2...n;给你一些点的坐标，让你输出所有格子含有i个点的个数有几个(i!=0)。

输入方式：第一行给你盒子有n个隔板有m个点，盒子的左上角(x1,y1)坐标和右下角坐标(x2,y2)。

接下来n行分别是隔板两端的x坐标，在接下来m行是点的坐标。

两题不同之处：POJ2398给的隔板顺序不同，需要排序；输出格式不同；输出内容不同；

POJ2318 AC代码：(红色部分是几何简单模板（参考算法竞赛大白书）)

#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#include<algorithm>#include<queue>using namespace std;<span style="color:#ff0000;">struct point{double x,y;point(double x=0,double y=0):x(x),y(y) { }};typedef point V;struct circle{point c;double r;circle(point c,double r):c(c),r(r){}point Point(double a){return point(c.x+cos(a)*r,+c.y+sin(a)*r);}};struct line{point c;V v;};V operator + (V a,V b){ return V(a.x+b.x,a.y+b.y);}V operator - (V a,V b){ return V(a.x-b.x,a.y-b.y);}V operator * (V a,double p){return V(a.x*p,a.y*p);}V operator / (V a,double p){return V(a.x/p,a.y/p);}bool operator < (const V& a,const V& b){    return a.x<b.x||(a.x==b.x&&a.y<b.y);}const double eps = 1e-10;int dcmp(double x){    if(fabs(x)<eps) return 0;    else        return x<0?-1:1;}bool operator == (const point& a,const point& b){return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;}double dot(V a,V b){return a.x*b.x+a.y*b.y; }double length(V a){return sqrt(dot(a,a));}double angle(V a,V b){return acos(dot(a,b)/length(a)/length(b)); }double cross(V a,V b){return a.x*b.y-a.y*b.x; }double area2(point a,point b,point c){return cross(b-a,c-a); }/*two line change point返回两线交点(保证有唯一交点)*/point twolineispoint(point p,V v,point q,V w){    V u=p-q;    double t=cross(w,u)/cross(v,w);    return p+v*t;}/*线段是否相交*/bool segmentmeet(point a1,point a2,point b1,point b2){    double c1=cross(a2-a1,b1-a1),c2=cross(a2-a1,b2-a1),           c3=cross(b2-b1,a1-b1),c4=cross(b2-b1,a1-b1);    return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;}/*点到直线距离*/point pointtoline(point p,point a,point b){    V v1=b-a,v2=p-a;    return fabs(cross(v1,v2))/length(v1);}</span>line a[5005];int ans[5005];int main(){    int n,m;    double strx,stry,endx,endy;    while(cin>>n&&n){        cin>>m>>strx>>stry>>endx>>endy;        memset(ans,0,sizeof(ans));        a[0].c.x=strx;        a[0].c.y=stry;        a[0].v.x=strx;        a[0].v.y=endy;        a[n+1].c.x=endx;        a[n+1].c.y=stry;        a[n+1].v.x=endx;        a[n+1].v.y=endy;        for(int i=1;i<=n;i++){            double xx,xx2;            cin>>xx>>xx2;            a[i].c.x=xx;            a[i].c.y=stry;            a[i].v.x=xx2;            a[i].v.y=endy;        }        for(int j=n+1;j<=n+m;j++){            point p;            cin>>p.x>>p.y;            for(int i=0;i<=n;i++){                if(dcmp((area2(p,a[i].c,a[i].v)))*dcmp((area2(p,a[i+1].c,a[i+1].v)))<0){                    ans[i]++;                    break;                }            }        }        for(int i=0;i<=n;i++){            cout<<i<<':'<<' '<<ans[i]<<endl;        }        cout<<endl;    }    return 0;}

POJ2398 AC代码：

#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#include<algorithm>#include<queue>using namespace std;<span style="color:#ff0000;">struct point{double x,y;point(double x=0,double y=0):x(x),y(y) { }};typedef point V;struct circle{point c;double r;circle(point c,double r):c(c),r(r){}point Point(double a){return point(c.x+cos(a)*r,+c.y+sin(a)*r);}};struct line{point c;V v;};V operator + (V a,V b){ return V(a.x+b.x,a.y+b.y);}V operator - (V a,V b){ return V(a.x-b.x,a.y-b.y);}V operator * (V a,double p){return V(a.x*p,a.y*p);}V operator / (V a,double p){return V(a.x/p,a.y/p);}bool operator < (const V& a,const V& b){    return a.x<b.x||(a.x==b.x&&a.y<b.y);}const double eps = 1e-10;int dcmp(double x){    if(fabs(x)<eps) return 0;    else        return x<0?-1:1;}bool operator == (const point& a,const point& b){return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;}double dot(V a,V b){return a.x*b.x+a.y*b.y; }double length(V a){return sqrt(dot(a,a));}double angle(V a,V b){return acos(dot(a,b)/length(a)/length(b)); }double cross(V a,V b){return a.x*b.y-a.y*b.x; }double area2(point a,point b,point c){return cross(b-a,c-a); }/*two line change point返回两线交点(保证有唯一交点)*/point twolineispoint(point p,V v,point q,V w){    V u=p-q;    double t=cross(w,u)/cross(v,w);    return p+v*t;}/*线段是否相交*/bool segmentmeet(point a1,point a2,point b1,point b2){    double c1=cross(a2-a1,b1-a1),c2=cross(a2-a1,b2-a1),           c3=cross(b2-b1,a1-b1),c4=cross(b2-b1,a1-b1);    return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;}/*点到直线距离*/point pointtoline(point p,point a,point b){    V v1=b-a,v2=p-a;    return fabs(cross(v1,v2))/length(v1);}</span>line a[5005];int ans[5005],anss[1005];int cmp(const line& a,const line& b){    return a.c<b.c||(a.c==b.c&&a.v<b.v);}int main(){    int n,m;    double strx,stry,endx,endy;    while(cin>>n&&n){        cin>>m>>strx>>stry>>endx>>endy;        memset(ans,0,sizeof(ans));        memset(anss,0,sizeof(anss));        a[0].c.x=strx;        a[0].c.y=stry;        a[0].v.x=strx;        a[0].v.y=endy;        a[n+1].c.x=endx;        a[n+1].c.y=stry;        a[n+1].v.x=endx;        a[n+1].v.y=endy;        for(int i=1;i<=n;i++){            double xx,xx2;            cin>>xx>>xx2;            a[i].c.x=xx;            a[i].c.y=stry;            a[i].v.x=xx2;            a[i].v.y=endy;        }        sort(a,a+n+1,cmp);        for(int j=n+1;j<=n+m;j++){            point p;            cin>>p.x>>p.y;            for(int i=0;i<=n;i++){                if(dcmp((area2(p,a[i].c,a[i].v)))*dcmp((area2(p,a[i+1].c,a[i+1].v)))<0){                    ans[i]++;                    break;                }            }        }        cout<<"Box\n";        for(int i=0;i<=n;i++){            if(ans[i]){                anss[ans[i]]++;            }        }        for(int i=0;i<=n;i++){            if(anss[i])            cout<<i<<':'<<' '<<anss[i]<<endl;        }    }    return 0;}