XJOI网上同步训练DAY1 T2

思路:似曾相识?...见http://www.cnblogs.com/qzqzgfy/p/5266874.html

一看时限还是4s!，于是就开开心心地打了70%的分，就是用容斥原理，就可以n^3解决问题了。

实际情况:10分，wtf

我的程序:

#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
const double eps=1e-6;
int tot[505][505],n,m,K;
const double Pi=acos(-1);
struct Point{
    double x,y,ang;
    int id,bel;
    Point(){}
    Point(double x0,double y0):x(x0),y(y0){}
}d[200005],f[200005],t[200005],p[200005];
struct Line{
    Point s,e;
    Line(){}
    Line(Point s0,Point e0):s(s0),e(e0){}
};
int sgn(double x){
    if (x<-eps)  return -1;
    if (x>eps) return 1;
    return 0;
}
bool cmp(Point p1,Point p2){
    return p1.ang<p2.ang;
}
double operator *(Point p1,Point p2){
    return p1.x*p2.y-p1.y*p2.x;
}
Point operator -(Point p1,Point p2){
    return Point(p1.x-p2.x,p1.y-p2.y);
}
int read(){
    char ch=getchar();int t=0,f=1;
    while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
    while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';ch=getchar();}
    return t*f;
}
bool inter(Line p1,Line p2){
    if (std::min(p1.s.x,p1.e.x)>std::max(p2.s.x,p2.e.x)||
        std::min(p1.s.y,p1.e.y)>std::max(p2.s.y,p2.e.y)||
        std::min(p2.s.x,p2.e.x)>std::max(p1.s.x,p1.e.x)||
        std::min(p2.s.y,p2.e.y)>std::max(p1.s.y,p1.e.y))
    return 0;
    double a,b,c,d;
    a=(p1.e-p1.s)*(p2.e-p1.s);
    b=(p1.e-p1.s)*(p2.s-p1.s);
    c=(p2.e-p2.s)*(p1.e-p2.s);
    d=(p2.e-p2.s)*(p1.s-p2.s);
    return (a*b<=eps)&&(c*d<=eps);
}
void sbpianfen1(){
    int ans=0;
    for (int i=1;i<=n;i++)
     for (int j=1;j<=m;j++)
      for (int k=1;k<=K;k++)
       for (int l=k+1;l<=K;l++)
        if (inter(Line(d[i],f[j]),Line(t[k],t[l]))) ans++;
    printf("%d
",ans);    
}
void updata(Point p1,Point p2,int v){
    if (p1.bel+p2.bel!=3) return;
    if (p1.bel>p2.bel) std::swap(p1,p2);
    tot[p1.id][p2.id]+=v;
}
void sbpianfen2(){
    for (int i=1;i<=K;i++){
        int cnt=0;
        for (int j=1;j<=n;j++){
            p[++cnt]=d[j];p[cnt].bel=1;
            p[cnt].ang=atan2(d[j].y-t[i].y,d[j].x-t[i].x);
        }
        for (int j=1;j<=m;j++){
            p[++cnt]=f[j];p[cnt].bel=2;
            p[cnt].ang=atan2(f[j].y-t[i].y,f[j].x-t[i].x);
        }
        for (int j=1;j<=K;j++){
            if (j==i) continue;
            p[++cnt]=t[j];p[cnt].bel=3;
            p[cnt].ang=atan2(t[j].y-t[i].y,t[j].x-t[i].x);
        }
        std::sort(p+1,p+1+cnt,cmp);
        for (int j=1;j<=cnt;j++){
            p[cnt+cnt+j]=p[cnt+j]=p[j];
            p[cnt+j].ang=p[j].ang+2*Pi;
            p[cnt+cnt+j].ang=p[cnt+j].ang+2*Pi;
        }
        int l=1;int numl=(p[1].bel==3);
        int numj=0,numr=0,numk=0;
        for (int j=1;j<=cnt;j++){
            if (p[j].bel==3) numj++;
            int r=j+1;numr=numj+(p[j+1].bel==3);
            while (sgn(p[l].ang-p[j].ang-Pi)<0) l++,numl+=(p[l].bel==3);
            for (int k=j+1,numk=numj+(p[j+1].bel==3);sgn(p[k].ang-p[j].ang-Pi)<=0;k++,numk+=(p[k].bel==3)){
                while (sgn(p[r+1].ang-p[k].ang-Pi)<=0) r++,numr+=(p[r].bel==3);
                updata(p[k],p[j],numk-numj-(numr-numl+(p[l].bel==3)));
            }
        }
    }
    int ans=0;
    for (int i=1;i<=n;i++)
     for (int j=1;j<=m;j++)
      ans+=tot[i][j];
    printf("%d
",ans/2);  
}
int main(){
    n=read();
    for (int i=1;i<=n;i++)
     d[i].x=read(),d[i].y=read(),d[i].id=i;
    m=read();
    for (int i=1;i<=m;i++)
     f[i].x=read(),f[i].y=read(),f[i].id=i;
    K=read();
    for (int i=1;i<=K;i++)
     t[i].x=read(),t[i].y=read(),t[i].id=i;  
    if (n<=100&&m<=100&&K<=100) {sbpianfen1();return 0;} 
    if (n<=500&&m<=500&&K<=500) {sbpianfen2();return 0;}
}

明明应该很科学啊。。

正解:n^2做法，枚举导弹发射井，然后极角排序，统计即可。

具体做法:枚举发射井作为原点，然后将其他类型的点排序，然后一边for走过去，ans就加等于在半平面内的基地数乘以导弹防御塔数，再减去在每个基地后边的导弹防御塔数之和，注意删掉相同极角的部分，这样就是答案了，果然我数学太差了，毛都不会。。。

#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#define ll long long
const double Pi=acos(-1);
ll ans=0;
int d[500005][2],n,K,m;
struct Point{
    int x,y;
    Point(){}
    Point(int x0,int y0):x(x0),y(y0){}
}S[200005],T[200005],E[200005];
struct node{
    long double w;
    Point p;
    int bz;
}Q[500005];
int read(){
    char ch=getchar();int t=0,f=1;
    while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
    while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';ch=getchar();}
    return t*f;
}
ll operator *(Point p1,Point p2){
    return (ll)p1.x*p2.y-(ll)p1.y*p2.x;
}
Point operator -(Point a,Point b){
    return Point(a.x-b.x,a.y-b.y);
}
bool cmp(const node a,const node b){
    return a.w<b.w||(!(a.p*b.p)&&a.bz<b.bz);
}
long double get(Point a){
    long double w=atan2(a.y,a.x);
    if (w<0) w=w+Pi+Pi;
    return w;
}
void calc(Point S){
    for (int i=1;i<=n;i++) Q[i].w=get(Q[i].p=E[i]-S),Q[i].bz=0;
    for (int i=1;i<=K;i++) Q[i+n].w=get(Q[i+n].p=T[i]-S),Q[i+n].bz=1;
    for (int i=1;i<=n+K;i++){
        Q[i+n+K]=Q[i];Q[i+n+K].w=Q[i].w+2*Pi;
        if (Q[i].bz) Q[i+n+K].bz=2;
    }
    std::sort(Q+1,Q+1+n+K+n+K,cmp);
    int sum=0,l=1,r=0,same=0;
    ll all=0,ans1=ans;
    for (int i=1;i<=n+K+n+K;i++){
        while (l<=r&&Q[d[l][0]].w+Pi<Q[i].w) all-=d[l++][1];
        if (Q[i].bz){
            if (i!=1&&Q[i].p*Q[i-1].p) same=0;
            ans+=(ll)(r-l+1)*sum-all;
            if (Q[i].bz==1) all+=sum-same,d[++r][0]=i,d[r][1]=sum-same; 
        }else{
            sum++;
            if (i!=1&&!(Q[i].p*Q[i-1].p)) same++;
            else same=1;
        }
    }
}
int main(){
    n=read();
    for (int i=1;i<=n;i++) E[i].x=read(),E[i].y=read();
    m=read();
    for (int i=1;i<=m;i++) S[i].x=read(),S[i].y=read();
    K=read();
    for (int i=1;i<=K;i++) T[i].x=read(),T[i].y=read();
    for (int i=1;i<=m;i++)
     calc(S[i]);
    printf("%lld
",ans); 
}