Protecting the Flowers

H - Protecting the Flowers

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit   Status   Practice   POJ 3262

Appoint description:   System Crawler  (2015-04-27)

Description

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 T_i minutes (1 ≤ T_i ≤ 2,000,000) away from its own barn. Furthermore, while waiting for transport, she destroysD_i (1 ≤ D_i ≤ 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 × T_i minutes (T_i to get there and T_i 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.

Input

Line 1: A single integer   N
Lines 2..   N+1: Each line contains two space-separated integers,   T_i  and   D_i, 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

Sample Output

86

Hint

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.

 

每头牛对应需要赶回的时间和吃草速度，求吃最少的草。

一个贪心的题，按速度除以时间排序；证明：取两头牛，其他的牛吃的草就可以看成常量，若先赶a牛，吃的草是2*t[a]*d[b],若是b牛，就是2*t[b]*d[a],就是比较这两个的大小。这样看就是一个水题了。

 

#include
#include
using namespace std;
struct Cow{
 long long t;
 long long d;
}cow[100005];
bool cmp(const Cow &a, const Cow &b)
{
 return a.d * b.t > b.d * a.t;
}
int main()
{
 ios::sync_with_stdio(false);
 long long m, n, i, j, sum, ans;
 cin >> n;
 sum = 0; ans = 0;
 for (i = 1; i <= n; i++)
 {
  cin >> cow[i].t >> cow[i].d;
 }
 for (i = 1; i <= n; i++)
  sum += cow[i].d;
 sort(cow + 1, cow + n+1 , cmp);
 for (i = 1; i <= n; i++)
 {
  sum -= cow[i].d;
  ans += sum * 2 * cow[i].t;
 }
 
 cout << ans << '\n';
 return 0;
}