HDU 4643 GSM

来源：互联网发布：运营数据分析专员编辑：程序博客网时间：2024/06/05 19:40

题意：有m个基站，手机只会接收距离最近的基站的信号，然后有k个询问，每次给你两个点a，b，问从a到b会更换几次基站

find(p)：返回离p点最近的基站的下标

如果find(p1)==find(p2)，则从p1到p2不会更换基站

如果p1，p2距离很小，则从p1到p2只会更换一次基站

否则二分求解

#include <cstdio>#include <cmath>const double eps = 1e-9;struct P{    double x, y;    P(){}    P(double _x, double _y){x = _x; y = _y;}};double dis2(P a, P b) {return (a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y);}int n, m;P city[55], base[55];int find(P a){    int k = 1; double d = dis2(a, base[1]);    for(int i = 2; i <= m; i++)    {        double nd = dis2(a, base[i]);        if(d > nd) k = i, d = nd;    }    return k;}int get_times(P l, P r){    int k1 = find(l), k2 = find(r);    if(k1 == k2) return 0;    if(dis2(l, r) < eps) return 1;    P mid((l.x + r.x)/2, (l.y + r.y)/2);    return get_times(l, mid) + get_times(mid, r);}int main(){    while(~scanf("%d%d", &n, &m))    {        for(int i = 1; i <= n; i++)            scanf("%lf%lf", &city[i].x, &city[i].y);        for(int i = 1; i <= m; i++)            scanf("%lf%lf", &base[i].x, &base[i].y);        int q, a, b;        scanf("%d", &q);        while(q--)        {            scanf("%d%d", &a, &b);            printf("%d\n", get_times(city[a], city[b]));        }    }    return 0;}

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