uvalive 4356

标签：uvalive

题意：有n个点在平面直接坐标线，给出了n个点坐标，然后问以(0，0)为圆心的扇形包含至少k个点最小面积。

题解：贪心，先把所有点按与x轴正半轴的角度排序，然后选出一个点当半径，枚举剩下点(半径小于第一个点)，更新最小面积值。

#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;
const int N = 5005;
const double PI = acos(-1.0);
struct P {
	int x, y, r;
	double angle;
}p[N];
int n, k;
double res, q[N];

bool cmp(P a, P b) {
	return a.angle < b.angle;
}

int main() {
	int cas = 1;
	while (scanf("%d%d", &n, &k) == 2 && n + k) {
		for (int i = 0; i < n; i++) {
			scanf("%d%d", &p[i].x, &p[i].y);
			p[i].angle = atan2(p[i].y, p[i].x);
			p[i].r = p[i].x * p[i].x + p[i].y * p[i].y;
		}
		if (k == 0) {
			printf("Case #%d: 0.00\n", cas++);
			continue;
		}
		sort(p, p + n, cmp);
		res = 1000000000.0;
		for (int i = 0; i < n; i++) {
			int num = 0;
			for (int j = 0; j < n; j++)
				if (p[j].r <= p[i].r)
					q[num++] = p[j].angle;
			if (num >= k) {
				double temp;
				for (int j = 0, l = k - 1; j < num; j++, l++) {
					if (l < num)
						temp = q[l] - q[j];
					else
						temp = q[l - num] - q[j] + PI * 2;
					res = min(res, temp * p[i].r / 2);
				}
			}
		}
		printf("Case #%d: %.2lf\n", cas++, res);
	}
	return 0;
}

uvalive 4356

标签：uvalive

原文地址：http://blog.csdn.net/hyczms/article/details/44733179