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.
题目的意思是指在花园里有一些奶牛在吃花朵，为了让损失达到最小，选择怎样送回奶牛；最后得出最少损失的花朵的数目； 题目给出奶牛的个数，每头奶牛送回去的时间和每分钟能吃掉花朵的个数；送回一头奶牛需要一来一回的时间。 这道题目利用了贪心的做法；重点是排序； 题目容易陷入先送每分钟吃的多的奶牛，然后送时间少的奶牛，但是如果吃的多，且送的时间也长的话，就得不偿失了； 所以题目真正的排序应该是先送(时间/吃的个数）最小的， 以两头奶牛想x,y举例，这样就只有一头奶牛在吃花朵，应该先送花朵数乘以另一头奶牛时间后得到的数值最大的奶牛 这是等待时间（同*2可以先省略 ）
这样被吃掉的花朵就会更少一点。 代码如下
#include<stdio.h> #include<string.h> #include<algorithm> #define ll long long using namespace std; struct w { ll a,b; }s[1010000]; bool cmp(w x,w y) { return (x.a*1.0/x.b)<(y.a*1.0/y.b); } int main() { int n; ll sum=0; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%lld %lld",&s[i].a,&s[i].b);//时间 花数 sum+=s[i].b; } sort(s,s+n,cmp); ll ans=0; for(int i=0;i<n;i++) { sum-=s[i].b; ans+=sum*s[i].a*2; } printf("%lld\n",ans); return 0; }