Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 11508 Accepted: 4671

Farmer John went to cut some wood and left N (2 ≤ N ≤ 100,000) cows eating the grass, as usual. When he returned, he found to his horror that the cluster of cows was in his garden eating his beautiful flowers. Wanting to minimize the subsequent damage, FJ decided to take immediate action and transport each cow back to its own barn.

Each cow i is at a location that is Ti minutes (1 ≤ Ti ≤ 2,000,000) away from its own barn. Furthermore, while waiting for transport, she destroys Di (1 ≤ Di ≤ 100) flowers per minute. No matter how hard he tries, FJ can only transport one cow at a time back to her barn. Moving cow i to its barn requires 2 × Ti minutes (Ti to get there and Ti to return). FJ starts at the flower patch, transports the cow to its barn, and then walks back to the flowers, taking no extra time to get to the next cow that needs transport.

Write a program to determine the order in which FJ should pick up the cows so that the total number of flowers destroyed is minimized.

Line 1: A single integer N Lines 2…N+1: Each line contains two space-separated integers, Ti and Di, that describe a single cow’s characteristics Output

Line 1: A single integer that is the minimum number of destroyed flowers Sample Input

6 3 1 2 5 2 3 3 2 4 1 1 6

86

FJ returns the cows in the following order: 6, 2, 3, 4, 1, 5. While he is transporting cow 6 to the barn, the others destroy 24 flowers; next he will take cow 2, losing 28 more of his beautiful flora. For the cows 3, 4, 1 he loses 16, 12, and 6 flowers respectively. When he picks cow 5 there are no more cows damaging the flowers, so the loss for that cow is zero. The total flowers lost this way is 24 + 28 + 16 + 12 + 6 = 86.

首先根据题意，一头牛的时间可以等效为 $2T$ ，接下来可以想一想几何意义，把第 $i$ 头牛想象成一个底为 $2T_{i}$ ，高为 $D_{i}$ 的矩形，总的花费应该是一系列高度不断下降的矩形在 $x≥0,y≥0$ 部分的积分面积，其中 $x$ 轴代表时间， $y$ 轴代表单位时间的花费。那么我们只需要构造积分面积最小的凹函数即可，只需让对角线斜率的绝对值大的矩形排在前面就可以了，也就是按照 $2T_{i}D_{i} $ 从大到小排序即可。

下面贴代码：

#include <cstdio> #include <algorithm> using namespace std; typedef long long ll; const int MAX_N = 100005; struct Cow { int T; int D; double x; bool operator< (const Cow& _Right)const { return x > _Right.x; } }cows[MAX_N]; int main() { int N; scanf("%d", &N); for (int i = 0; i < N; ++i) { Cow& cur = cows[i]; scanf("%d%d", &cur.T, &cur.D); cur.T <<= 1; cur.x = (double)cur.D / cur.T; } sort(cows, cows + N); ll ans = 0, t = 0; for (int i = 0; i < N; ++i) { const Cow& cur = cows[i]; ans += cur.D * t; t += cur.T; } printf("%lld\n", ans); return 0; }
转载请注明原文地址: https://win8.8miu.com/read-900015.html