SGU114 Telecasting station
题目大意
有N个城市分布在x轴上,每个城市有一定数量的人口
现在要修一座电视塔,而每个城市的不满意度为人口数乘以该城市到电视塔的距离
问在哪里修电视塔,使得所有城市总的不满意度最低
算法思路
将城市按照位置排序后,电视塔修在左右人口都少于一半的地方
时间复杂度: O(NlogN)
代码
/**
* Copyright (c) 2015 Authors. All rights reserved.
*
* FileName: 114.cpp
* Author: Beiyu Li <sysulby@gmail.com>
* Date: 2015-05-22
*/
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
typedef long long LL;
typedef pair<int, int> Pii;
const int inf = 0x3f3f3f3f;
const LL infLL = 0x3f3f3f3f3f3f3f3fLL;
const int maxn = 15000 + 5;
int n;
Pii a[maxn];
int main()
{
int tot = 0, now = 0;
scanf("%d", &n);
rep(i,n) {
scanf("%d%d", &a[i].first, &a[i].second);
tot += a[i].second;
}
sort(a, a + n);
rep(i,n) if ((now += a[i].second) * 2 >= tot) {
printf("%d.00000\n", a[i].first);
break;
}
return 0;
}