[hdu4631 Sad Love Story]最近点对，枚举

题意：S是平面内点的集合，初始为空，每次向集合里面加入一个点P(x,y)，询问S内最近点对的距离的平方和

思路：设当前集合的答案为D，则找到集合里面横坐标在(x-√D,x+√D)内的数，用它们来更新答案，一边更新答案一边还要更新右边界x+√D，此时的更新注意不要用浮点数开平方算具体右边界，改用判断即可，否则超时。

 

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#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define X                   first
#define Y                   second
#define pb                  push_back
#define mp                  make_pair
#define all(a)              (a).begin(), (a).end()
#define fillchar(a, x)      memset(a, x, sizeof(a))
#define copy(a, b)          memcpy(a, b, sizeof(a))

typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;

//#ifndef ONLINE_JUDGE
void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);}
void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>
void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1;
while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
//#endif
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}
template<typename T>
void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];}
template<typename T>
void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];}

const double PI = acos(-1.0);
const int INF = 1e9 + 7;
const double EPS = 1e-8;

/* -------------------------------------------------------------------------------- */

multiset<pii> S;

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
    int T, n, ax, bx, cx, ay, by, cy, x, y;
    cin >> T;
    while (T --) {
        scanf("%d%d%d%d%d%d%d", &n, &ax, &bx, &cx, &ay, &by, &cy);
        S.clear();
        x = bx % cx;
        y = by % cy;
        S.insert(mp(x, y));
        ll ans = 0, mind = (ll)INF * INF;
        for (int i = 1; i < n; i ++) {
            x = ((ll)x * ax + bx) % cx;
            y = ((ll)y * ay + by) % cy;
            int dx = (int)(sqrt(mind * 1.0) + 1);
            multiset<pii>::iterator iter = S.lower_bound(mp(x - dx, -INF));
            for (; iter != S.end(); iter ++) {
                ll X = (*iter).X - x, Y = (*iter).Y - y;
                ll buf = X * X + Y * Y;
                if (X >= 0 && X * X >= mind) break;
                umin(mind, buf);
            }
            ans += mind;
            S.insert(mp(x, y));
        }
        cout << ans << endl;
    }
    return 0;
}
posted @ 2015-08-12 15:52  jklongint  阅读(225)  评论(0)  编辑  收藏  举报