HDU4631 Sad Love Story（最近点对）转自网络

转自http://blog.youkuaiyun.com/acm_zl/article/details/9634443

这道题目的意思就是不断加入n个点。当点数>=2的时候，每加入一个点累加两点间最近距离的平方。按照给定的Ax,Bx,Cx,Ay,By,Cy,以及递推式可以产生n个点。
所有首先n个点，做一下最近点对，复杂度O(nlogn)然后产生的最近点对，对于编号在最近点对后面的结果都可以累加了，同时后面的点也不需要了。所有去掉一部分点再次进行最近点对。这样不断重复，直到剩下一个点为止。

#include <stdio.h>  
#include <algorithm>  
#include <string.h>  
#include <iostream>  
#include <map>  
#include <vector>  
#include <queue>  
#include <set>  
#include <string>  
#include <math.h>  
using namespace std;  
const int MAXN = 500010;  
struct Point  
{  
    int x,y;  
    int id;  
    int index;  
    Point(int _x = 0,int _y = 0,int _index = 0)  
    {  
        x = _x;  
        y = _y;  
        index = _index;  
    }  
};  
Point P[MAXN];  
long long dist(Point a,Point b)  
{  
    return ((long long)(a.x-b.x)*(a.x-b.x) + (long long)(a.y-b.y)*(a.y-b.y));  
}  
Point p[MAXN];  
Point tmpt[MAXN];  
bool cmpxy(Point a,Point b)  
{  
    if(a.x != b.x)return a.x < b.x;  
    else return a.y < b.y;  
}  
bool cmpy(Point a,Point b)  
{  
    return a.y < b.y;  
}  
pair<int,int> Closest_Pair(int left,int right)  
{  
    long long d = (1LL<<50);  
    if(left == right)return make_pair(left,right);  
    if(left + 1 == right)  
        return make_pair(left,right);  
    int mid = (left+right)/2;  
    pair<int,int>pr1 = Closest_Pair(left,mid);  
    pair<int,int>pr2 = Closest_Pair(mid+1,right);  
    long long d1,d2;  
    if(pr1.first == pr1.second)  
        d1 = (1LL<<50);  
    else d1 = dist(p[pr1.first],p[pr1.second]);  
    if(pr2.first == pr2.second)  
        d2 = (1LL<<50);  
    else d2 = dist(p[pr2.first],p[pr2.second]);  
    pair<int,int>ans;  
    if(d1 < d2)ans = pr1;  
    else ans = pr2;  
    d = min(d1,d2);  
    int k = 0;  
    for(int i = left;i <= right;i++)  
    {  
        if((long long)(p[mid].x - p[i].x)*(p[mid].x - p[i].x) <= d)  
            tmpt[k++] = p[i];  
    }  
    sort(tmpt,tmpt+k,cmpy);  
    for(int i = 0;i <k;i++)  
    {  
        for(int j = i+1;j < k && (long long)(tmpt[j].y - tmpt[i].y)*(tmpt[j].y - tmpt[i].y) < d;j++)  
        {  
            if(d > dist(tmpt[i],tmpt[j]))  
            {  
                d = dist(tmpt[i],tmpt[j]);  
                ans = make_pair(tmpt[i].id,tmpt[j].id);  
            }  
        }  
    }  
    return ans;  
}  
int main()  
{   
    int T;  
    int n,Ax,Ay,Bx,By,Cx,Cy;  
    scanf("%d",&T);  
    while(T--)  
    {  
        scanf("%d%d%d%d%d%d%d",&n,&Ax,&Bx,&Cx,&Ay,&By,&Cy);  
        P[0] = Point(0,0,0);  
        for(int i = 1;i <= n;i++)  
        {  
            long long x= ((long long)P[i-1].x*Ax + Bx)%Cx;  
            long long y = ((long long)P[i-1].y*Ay + By)%Cy;  
            P[i] = Point(x,y,i);  
        }  
        int end = n;  
        long long ans = 0;  
        while(end > 1)  
        {  
            for(int i = 0;i < end;i++)  
                p[i] = P[i+1];  
            sort(p,p+end,cmpxy);  
            for(int i = 0;i < end;i++)  
                p[i].id = i;  
            pair<int,int>pr = Closest_Pair(0,end-1);  
            int Max = max(p[pr.first].index,p[pr.second].index);  
            ans += (long long)(end-Max+1)*dist(p[pr.first],p[pr.second]);  
            end = Max-1;  
        }  
        printf("%I64d\n",ans);  
    }  
    return 0;  
}