POJ - 3714 分治

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题意：

给出两个集合，每个集合中有n个点，求属于不同集合的两个点之间的最短距离。

思路：

分治，套用最接近点对问题的方法，只要在保存res的时候判断是否是属于同一个集合即可。

代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;typedef long long ll;const int MAXN = 2e5 + 10;const ll INF = 0x3f3f3f3f3f3f3f3f;struct Point {    double x, y;    int flag;}p[MAXN], q[MAXN];bool cmpx(const Point &a, const Point &b) {    return a.x < b.x;}bool cmpy(const Point &a, const Point &b) {    return a.y < b.y;}double dist(Point a, Point b) {    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));}double solve(int l, int r) {    if (l == r) return INF;    if (l + 1 == r) {        if (p[l].flag != p[r].flag) return dist(p[l], p[r]);        return INF;    }    int m = (l + r) >> 1;    double res = min(solve(l, m), solve(m + 1, r));    int cnt = 0;    for (int i = l; i <= r; i++)        if (fabs(p[i].x - p[m].y) <= res) q[++cnt] = p[i];    sort (q + 1, q + cnt + 1, cmpy);    for (int i = 1; i <= cnt; i++) {        for (int j = i + 1; j <= cnt; j++) {            if (q[j].y - q[i].y >= res) break;            if (q[i].flag != q[j].flag)                res = min(res, dist(q[i], q[j]));        }    }    return res;}int main() {    int T;    scanf("%d", &T);    while (T--) {        int n;        scanf("%d", &n);        for (int i = 1; i <= n; i++) {            scanf("%lf%lf", &p[i].x, &p[i].y);            p[i].flag = 0;        }        for (int i = n + 1; i <= 2 * n; i++) {            scanf("%lf%lf", &p[i].x, &p[i].y);            p[i].flag = 1;        }        sort (p + 1, p + 1 + 2 * n, cmpx);        printf("%.3f\n", solve(1, 2 * n));    }    return 0;}

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