SGI114 Telecasting station

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 <[e-mail]> * 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;}

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