Protecting the Flowers
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 4237 Accepted: 1710
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 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.
Input
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
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.
Source
USACO 2007 January Silver
题意：
有n个牛在FJ的花园乱吃。
所以FJ要赶他们回牛棚。
每个牛在被赶走之前每秒吃Di个花朵。赶它回去FJ来回要花的总时间是Ti×2。在被赶走的过程中，被赶走的牛就不能乱吃
思路：
贪心策略，对牛进行排序，排序的标准是，假设牛A与牛B要选一头赶走，我们首先要选择破坏最大的一头牛赶走，留破坏小
的牛。他们的破坏着呢麽计算呢？假设先赶走牛A，那么牛B造成的损失是2×TA×DB，先赶走牛B，那么牛A造成的损失是2×TA×DB，
所以，只要判断TA×DB与TA×DB谁大，就知道该先赶走谁了，所以数组排序的标准就是---Ti×Dj>Tj×Di
AC代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define INF 0x11111110
using namespace std;
struct node
{
__int64 x;
__int64 y;
}a[100010];
int cmp(node n,node m)
{
return n.y*m.x>m.y*n.x;
}
int main()
{
int i,j,n;
__int64 sum,num;
while(scanf("%d",&n)!=EOF)
{
sum=0;
memset(a,0,sizeof(a));
for(i=0;i<n;i++)
{
scanf("%I64d %I64d",&a[i].x,&a[i].y);
sum+=a[i].y;
}
sort(a,a+n,cmp);
num=0;
for(i=0;i<n;i++)
{
sum-=a[i].y;
num+=sum*(a[i].x*2);
}
printf("%I64d\n",num);
}
return 0;
}