UVA - 10245 The Closest Pair Problem

Posted 2021-01-21 widsom

UVA - 10245

思路:

平面分治

inplace_merge()可以用来归并排序

代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 1e4 + 5;
const int INF = 0x3f3f3f3f;
pair<double, double> p[N];
bool cmp(pair<double, double> a, pair<double, double> b) {
    return a.se < b.se;
}
double solve(pair<double, double> *a, int n) {
    if(n <= 1) return INF;
    int m = n/2;
    double mx = a[m].fi;
    double d = min(solve(a, m), solve(a+m, n-m));
    inplace_merge(a+1, a+m+1, a+n+1, cmp);
    vector<pair<double, double> >vc;
    for (int i = 1; i <= n; i++) {
        if(abs(a[i].fi - mx) > d) continue;
        for (int j = 0; j < vc.size(); j++) {
            double dx = abs(a[i].fi - vc[vc.size()-1-j].fi);
            double dy = abs(a[i].se - vc[vc.size()-1-j].se);
            if(dy > d) break;
            d = min(d, sqrt(dx*dx + dy*dy));
        }
        vc.pb(a[i]);
    }
    return d;
}
int main() {
    int n;
    while(~scanf("%d", &n) && n) {
        for (int i = 1; i <= n; i++) scanf("%lf %lf", &p[i].fi, &p[i].se);
        sort(p+1, p+n+1);
        double ans = solve(p, n);
        if(ans < 10000) printf("%.4f
", ans);
        else printf("INFINITY
");
    }
    return 0;
}