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+10;
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=0, yy=0;
    for( int i=0; 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 = 0;
    st.insert(p[0]);
    LL mi = 1LL<<60;
    for (int i = 1; i < N; i++){
        Point t = p[i];
        st.insert(t);
        it1 = it2 = it3 = st.lower_bound(t);
        int k = 5;
        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 == 0) break;
    }
    printf("%I64d
",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 0;
}