bzoj 1634: [Usaco2007 Jan]Protecting the Flowers 护花
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 cows were in his garden eating his beautiful flowers. Wanting to minimize the subsequent damage, FJ decided to take immediate action and transport the cows back to their barn. Each cow i is at a location that is Ti minutes (1 <= Ti <= 2,000,000) away from the 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 the barn. Moving cow i to the barn requires 2*Ti minutes (Ti to get there and Ti to return). 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
第1行输入N,之后N行每行输入两个整数Ti和Di.
Output
* Line 1: A single integer that is the minimum number of destroyed flowers
一个整数,表示最小数量的花朵被吞食.
Sample Input
3 1
2 5
2 3
3 2
4 1
1 6
Sample Output
HINT
约翰用6,2,3,4,1,5的顺序来运送他的奶牛.
解:贪心,
可以发现,调换序列中a与b的位置对于a和b以外的数是没有任何影响的
如果 x.t*y.d<y.t*x.d(这是关键)
x排在y前面会比y排在x前面更优,所以可以按照这个规则排序模拟
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cstdio> #include<cstring> #include<algorithm> #define ll long long using std::sort; const int N=100010; struct node { int t,d; }map[N]; ll ans=0,sum=0; bool cmp(node x,node y){return x.t*y.d<y.t*x.d;} int main() { int n; scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%d %d",&map[i].t,&map[i].d); sum+=map[i].d; } sort(map+1,map+1+n,cmp); for(int i=1;i<=n;i++) { sum-=map[i].d; ans+=sum*(map[i].t<<1); } printf("%lld",ans); return 0; }