HDU 5517 题解

题意：给出多重集合A，B。A为两个数的有序对多重集合，B为三个数的有序对多重集合，A有n个元素(a_i,b_i)，B有m个元素(c_i,d_i,e_i)(依照多重集合定义，元素可以重复)，定义集合C为所有有序对(a,c,d)的集合，其中(a,b)∈A，(c,d,e)∈B，且b=e

TOP(C)为所有有序对（a,b,c）∈C的集合，其中（a,b,c）满足C中不存在一个有序对(d,e,f),使得d>=a && e>=b && f>=c.求TOP(C)的元素个数

1<=N<=100000，1<=m<=100000

1<=a_i,b_i<=100000;1<=c_i,d_i<=1000;1=<e_i<=100000;

共1~10组数据，6000MS

算法/思路：树状数组：直接构造C集合，但其中的a满足a=max{a_i，(a_i,b_i)∈A}，然后按a从大到小考虑每个a对应的有序对(b,c)，当前有序对满足条件当且仅当对所有曾经扫描过的(b_i,c_i)，有b>b_i或c>c_i,该过程可用树状数组维护(对于所有有序对(b,c),用log(m)时间求出b大于b_i时，c的最小值)。时间复杂度O(n+mlogm)

 

实现细节：由于A，B是多重集合，对于a要加计数器，(b,c)重复考虑即可，树状数组的实现，由于要维护区间[b,maxn]，所以把所有位置i变成maxn-i+1丢进去。

 

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

#define rep(i,a,b) for (int i=a;i<=b;++i)

const double eps=1e-7;

struct point{
    double x,y;
    point(){}
    point (double a,double b): x(a),y(b) {}

    friend point operator + (const point &a,const point &b){
        return point(a.x+b.x,a.y+b.y);
    }

    friend point operator - (const point &a,const point &b){
        return point(a.x-b.x,a.y-b.y);
    }

    friend point operator * (const double &r,const point &a){
        return point(r*a.x,r*a.y);
    }

    friend bool operator == (const point &a,const point &b){
        return (abs(a.x-b.x)<eps && abs(a.y-b.y)<eps);
    }

    double norm(){
        return sqrt(x*x+y*y);
    }
};

inline double det(point a,point b) {return a.x*b.y-a.y*b.x;}

inline bool line_cross_segment(point s,point t,point a,point b)
{
    return det(s-a,t-a)*det(s-b,t-b)<-eps;
}

int n,tot,x,y;
point s[50],t[50],goal;
int ans,temp;

int check(point a,point b)
{
    int temp=0;
    rep(i,1,n)
        if (line_cross_segment(s[i],t[i],a,b)) ++temp;
    return temp;
}

int main()
{
    scanf("%d",&n);
    rep(i,1,n) scanf("%lf%lf%lf%lf",&s[i].x,&s[i].y,&t[i].x,&t[i].y);
    ans=100000;
    scanf("%lf%lf",&goal.x,&goal.y);
    rep(i,1,n)
    {
        ans=min(ans,check(s[i],goal));
        ans=min(ans,check(t[i],goal));
    }
    if (n==0) ans=0;
    printf("Number of doors = %d\n",ans+1);
    return 0;
}