uva 10245(分治)

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题意：一些点的坐标给出，求出这些点最近的距离，大于10000输出INF...

题解：直接遍历超时。。于是学会了用分治，先把坐标按x的大小升序排列，然后平均分成两部分，递归找出左半边点最小的距离d1，右半边点最小的距离d2，那么很明显，最后最小的距离肯定是，d1 或者 d2 或者中间的左边某个点和右边某个点的连线d3，这三者中最短的一个。然后开始找跨中间的两点最小距离，只要左边点与中间点的x坐标差或右边点与中间点的x坐标差小于min(d1, d2)，就计算与之比较选出最小值。

#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std;const int N = 10005;struct Point {double x, y;}p[N];int cmpx(Point a, Point b) {if (a.x != b.x)return a.x < b.x;return a.y < b.y;}double count(Point p1, Point p2) {return sqrt((p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y));}double solve(int l, int r) {if (l == r)return 100000.0;if (r - l == 1)return count(p[l], p[r]);double temp1 = solve(l, (l + r) / 2);double temp2 = solve((l + r) / 2, r);double minn = temp1 > temp2 ? temp2 : temp1;int mid = (l + r) / 2;for (int i = mid - 1; i >= l && p[mid].x - p[i].x < minn; i--)for (int j = mid + 1; j <= r && p[j].x - p[mid].x < minn; j++) {double temp = count(p[i], p[j]);if (temp < minn)minn = temp;}return minn;}int main() {int n;double mindist;while (scanf("%d", &n) && n) {for (int i = 0; i < n; i++)scanf("%lf%lf", &p[i].x, &p[i].y);sort(p, p + n, cmpx);mindist = solve(0, n - 1);if (mindist > 10000)printf("INFINITY\n");elseprintf("%.4lf\n", mindist);}return 0;}

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