hdu4631Sad Love Story(多校3）（最接近点对）

http://acm.hdu.edu.cn/showproblem.php?pid=4631

比赛的时候搜到了最接近点对的求法 Nlog(N） 又估摸着依次插入求的话会TLE 想了想觉得可以先把最近的位置求出来 然后后面的直接不用求了 依次直到减完 又觉得可能会有变态的数据每次最近的都在最后面 没敢写。。后来 发现它出现在题解的方法三中。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define  N 500005
#define LL long long
#define INF  1000000000000000000
int mm;
LL d;
struct Point
{
    LL x;
    LL y;
    int p;
}point[N],tt[N];
int tmpt[N];

bool cmpxy(const Point& a, const Point& b)
{
    if(a.x != b.x)
        return a.x < b.x;
    return a.y < b.y;
}
bool cmpy(const int& a, const int& b)
{
    return point[a].y < point[b].y;
}

LL min(LL a, LL b)
{
    return a < b ? a : b;
}

LL dis(int i, int j)
{
    return (point[i].x-point[j].x)*(point[i].x-point[j].x)
                + (point[i].y-point[j].y)*(point[i].y-point[j].y);
}
void Closest_Pair(int left, int right)
{
    if(left==right)
    {
        return ;
    }
    if(left + 1 == right)
    {
         LL xx = dis(left, right);
         if(d>xx)
         {

             d = xx;
             mm = max(point[left].p,point[right].p);
         }
         return ;
     }
    int mid = (left+right)>>1;
    Closest_Pair(left,mid);
    Closest_Pair(mid+1,right);
    int i,j,k=0;
    for(i = left; i <= right; i++)
    {
        if((point[mid].x-point[i].x)*(point[mid].x-point[i].x) <= d)
            tmpt[k++] = i;
    }
    sort(tmpt,tmpt+k,cmpy);
    for(i = 0; i < k; i++)
    {
        for(j = i+1; j < k && (point[tmpt[j]].y-point[tmpt[i]].y)*(point[tmpt[j]].y-point[tmpt[i]].y)<d; j++)
        {
            LL d3 = dis(tmpt[i],tmpt[j]);
            if(d > d3)
            {
                 d = d3;
                 mm = max(point[tmpt[j]].p,point[tmpt[i]].p);
            }
        }
    }
}
int main()
{
    int t,a[10],i;
    cin>>t;
    while(t--)
    {
        for(i = 1; i <= 7 ; i++)
        cin>>a[i];
        point[i].x = 0;
        point[i].y = 0;
        for(i = 1; i <= a[1] ; i++)
        {
            point[i].x = (point[i-1].x*a[2]+a[3])%a[4];
            point[i].y = (point[i-1].y*a[5]+a[6])%a[7];
            point[i].p = i;
            tt[i].x = point[i].x;
            tt[i].y = point[i].y;
            tt[i].p = i;
        }
        LL s=0;
        int n = a[1];
        while(1)
        {
            sort(point+1,point+n+1,cmpxy);
            d = INF;
            Closest_Pair(1,n);
            s+=(n-mm+1)*d;
            n = mm-1;
            for(i = 1 ; i <= n ; i++)
            {
                point[i].x = tt[i].x;
                point[i].y = tt[i].y;
                point[i].p = tt[i].p;
            }
            if(n<=1)
            break;
        }
        cout<<s<<endl;
    }
    return 0;
}