HDU 4631 Sad Love Story 平面内最近点对

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

题意: 在平面内依次加点,求每次加点后最近点对距离平方的和

因为是找平面最近点对...所以加点以后这个最短距离一定是递减的...所以最后会形成这样一个函数图像

所以我们只要从后往前依次删点即可...

15秒惊险水过...不过我最小点对的木板肯定写挂了,卡时限的话估计过不了...

请用G++交...C++会TLE...当然我也无法解释这个问题...估计是我傻逼

 

/********************* Template ************************/
#include <set>
#include <map>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;#define EPS         1e-8
#define MAXN        (int)5e5+5
#define MOD         (int)1e9+7
#define PI          acos(-1.0)
#define INF         ((1LL)<<50)
#define max(a,b)    ((a) > (b) ? (a) : (b))
#define min(a,b)    ((a) < (b) ? (a) : (b))
#define max3(a,b,c) (max(max(a,b),c))
#define min3(a,b,c) (min(min(a,b),c))
#define BUG         cout<<"BUG! "<<endl
#define LINE        cout<<"------------------"<<endl
#define L(t)        (t << 1)
#define R(t)        (t << 1 | 1)
#define Mid(a,b)    ((a + b) >> 1)
#define lowbit(a)   (a & -a)
#define FIN         freopen("out.txt","w",stdout)
#pragma comment     (linker,"/STACK:102400000,102400000")// typedef long long LL;
// typedef unsigned long long ULL;
typedef __int64 LL;
// typedef unisigned __int64 ULL;
// int gcd(int a,int b){ return b?gcd(b,a%b):a; }
// int lcm(int a,int b){ return a*b/gcd(a,b); }/*********************   F   ************************/
struct point
{LL x,y;int pos;int id;point(double a = 0,double b = 0,int c = 0){x = a ; y = b ; pos = c;}
}p[MAXN],t[MAXN],tmp[MAXN];LL n,ax,bx,cx,ay,by,cy;bool cmp(point a,point b){if(a.x == b.x) return a.y < b.y;return a.x < b.x;
}bool cmp1(point a,point b){return a.y < b.y;
}LL dist(point a ,point b){return (a.x-b.x) * (a.x-b.x) + (a.y-b.y) * (a.y-b.y);
}/**  二维空间找最近点对*  返回排序后点位置的pair<int,int>*/
pair<int,int> Closest_Pair(int l ,int r){if(l == r || l+1 == r) return make_pair(l,r); //1个点,2个点 直接return;int m = Mid(l,r);       // (l+r)/2pair<int,int> dl = Closest_Pair(l,m);pair<int,int> dr = Closest_Pair(m+1,r);LL ldis,rdis;       //左部分的最值 右部分的最值LL ans_dis;         //左中右三部分最值if(dl.first == dl.second) ldis = INF;   //判重else ldis = dist(p[dl.first],p[dl.second]);if(dr.first == dr.second) rdis = INF;else rdis = dist(p[dr.first],p[dr.second]);pair<int,int> ans = ldis < rdis ? dl : dr ; //左右两部分的最值点对ans_dis = min(ldis,rdis);                   //左右两部分的最值// 从中向左右两边找在[p[m].x-d,p[m].x+d]的平面内所有点// 这以后的复杂度就不太好估计了...// 这段模板是用暴力找的...我只做了一点点优化...但为什么加剪枝时间还多了这我不太理解囧int cnt = 0;
//    for(int i = l; i <= r; i++)
//    {
//        if((long long)(p[m].x - p[i].x)*(p[m].x - p[i].x) <= ans_dis)
//            tmp[cnt++] = p[i];
//    }for(int i = m ; i >= l ; i--){LL q = (p[m].x - p[i].x) * (p[m].x - p[i].x);if(p[i].x < p[m].x - q) break;if(q <= ans_dis){tmp[cnt++] = p[i];}}for(int i = m+1 ; i <= r ; i++){LL q = (p[m].x - p[i].x) * (p[m].x - p[i].x);if(p[i].x > p[m].x + q) break;if(q <= ans_dis){tmp[cnt++] = p[i];}}//按y方向进行筛选 ,相隔大于d的点可以直接跳过sort(tmp,tmp+cnt,cmp1);for(int i = 0 ; i < cnt ; i++){for(int j = i+1 ; j < cnt ; j++){if((tmp[i].y - tmp[j].y) * (tmp[i].y - tmp[j].y) >= ans_dis)break;if(dist(tmp[i],tmp[j]) < ans_dis){ans_dis = dist(tmp[i],tmp[j]);ans = make_pair(tmp[i].id,tmp[j].id);}}}return ans;
}void pre(){t[0].x = bx % cx;t[0].y = by % cy;t[0].pos = 0;for(int i = 1 ; i <= n ; i++) {t[i].x = (t[i-1].x * ax + bx) % cx;t[i].y = (t[i-1].y * ay + by) % cy;t[i].pos = i;}
}
int main()
{//FIN;int T;scanf("%d",&T);while(T--){scanf("%d%I64d%I64d%I64d%I64d%I64d%I64d",&n,&ax,&bx,&cx,&ay,&by,&cy);pre();LL res  = 0 ;while(n){for(int i = 0 ; i < n ; i++)p[i] = t[i];sort(p,p+n,cmp);for(int i = 0 ; i < n ; i++)p[i].id = i;pair<int,int> ans = Closest_Pair(0,n-1);int last = max(p[ans.first].pos,p[ans.second].pos);res += (dist(p[ans.first],p[ans.second]) * (n - last));n = last ;}printf("%I64d\n",res);}return 0;
}