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~
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <algorithm> 5 #include <vector> 6 #include <set> 7 using namespace std; 8 const int MAX=5e5+10; 9 typedef long long LL; 10 LL ax, bx, cx, ay, by, cy; 11 int T, N; 12 struct Point 13 { 14 LL x, y; 15 Point(){} 16 Point(LL a, LL b):x(a), y(b){}; 17 bool operator < (const Point &p)const{ 18 return x < p.x || (x == p.x && y < p.y); 19 } 20 21 }p[MAX]; 22 LL sqr(LL x) 23 { 24 return x*x; 25 } 26 LL Dis( Point A, Point B ) 27 { 28 return sqr(A.x-B.x)+sqr(A.y-B.y); 29 } 30 31 void Init( ) 32 { 33 LL xx=0, yy=0; 34 for( int i=0; i<N; ++ i ){ 35 xx=(xx*ax+bx)%cx; 36 yy=(yy*ay+by)%cy; 37 p[i]=Point( xx, yy ); 38 } 39 } 40 multiset<Point>st; 41 multiset<Point>::iterator it1, it2, it3; 42 void gao(){ 43 st.clear(); 44 LL ans = 0; 45 st.insert(p[0]); 46 LL mi = 1LL<<60; 47 for (int i = 1; i < N; i++){ 48 Point t = p[i]; 49 st.insert(t); 50 it1 = it2 = it3 = st.lower_bound(t); 51 int k = 5; 52 while (k--){ 53 if (it1 != st.begin()) it1--; 54 if (it3 != it1){ 55 LL d1 = Dis(t,*it1); 56 if (d1<mi){ 57 mi = d1; 58 } 59 } 60 if (it2 != st.end()) it2++; 61 if (it2 != st.end() && it2!= it3){ 62 LL d2 = Dis(t,*it2); 63 if (d2 < mi){ 64 mi = d2; 65 } 66 } 67 } 68 ans += mi; 69 if (mi == 0) break; 70 } 71 printf("%I64d ",ans); 72 } 73 int main() 74 { 75 scanf("%d", &T); 76 while(T--){ 77 scanf("%d", &N); 78 scanf("%I64d%I64d%I64d%I64d%I64d%I64d", &ax, &bx, &cx, &ay, &by, &cy); 79 Init(); 80 gao(); 81 } 82 return 0; 83 }