hdu 4631Sad Love Story<计算几何>

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

题意：依次给你n个点，每次求出当前点中的最近点对，输出所有最近点对的和；

思路：按照x排序，然后用set维护，每次插入只更新当前点和插入点前后几个位置~

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <set>
 using namespace std;
 const int MAX=5e5+;
 typedef long long LL;
 LL ax, bx, cx, ay, by, cy;
 int T, N;
 struct Point
 {
     LL x, y;
     Point(){}
     Point(LL a, LL b):x(a), y(b){};
    bool operator < (const Point &p)const{
         return x < p.x || (x == p.x && y < p.y);
     }

 }p[MAX];
 LL sqr(LL x)
 {
     return x*x;
 }
 LL Dis( Point A, Point B )
 {
     return sqr(A.x-B.x)+sqr(A.y-B.y);
 }

 void Init( )
 {
     LL xx=, yy=;
     for( int i=; i<N; ++ i ){
         xx=(xx*ax+bx)%cx;
         yy=(yy*ay+by)%cy;
         p[i]=Point( xx, yy );
     }
 }
 multiset<Point>st;
 multiset<Point>::iterator it1, it2, it3;
 void gao(){
     st.clear();
     LL ans = ;
     st.insert(p[]);
     LL mi = 1LL<<;
     for (int i = ; i < N; i++){
         Point t = p[i];
         st.insert(t);
         it1 = it2 = it3 = st.lower_bound(t);
         int k = ;
         while (k--){
             if (it1 != st.begin()) it1--;
             if (it3 != it1){
                 LL d1 = Dis(t,*it1);
                 if (d1<mi){
                     mi = d1;
                 }
             }
             if (it2 != st.end()) it2++;
             if (it2 != st.end() && it2!= it3){
                 LL d2 = Dis(t,*it2);
                 if (d2 < mi){
                     mi = d2;
                 }
             }
         }
         ans += mi;
         if (mi == ) break;
     }
     printf("%I64d\n",ans);
 }
 int main()
 {
     scanf("%d", &T);
     while(T--){
         scanf("%d", &N);
         scanf("%I64d%I64d%I64d%I64d%I64d%I64d", &ax, &bx, &cx, &ay, &by, &cy);
         Init();
         gao();
     }
     return ;
 }