BZOJ1634: [Usaco2007 Jan]Protecting the Flowers 護花

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.

   約翰留下他的N只奶牛上山採木．他離開的時候，她們像往常一樣悠閒地在草場裏喫草．可是，當他回來的時候，他看到了一幕慘劇：牛們正躲在他的花園裏，啃食着他心愛的美麗花朵！爲了使接下來花朵的損失最小，約翰趕緊採取行動，把牛們送回牛棚． 牛們從1到N編號．第i只牛所在的位置距離牛棚Ti(1≤Ti《2000000)分鐘的路程，而在約翰開始送她回牛棚之前，她每分鐘會啃食Di(1≤Di≤100)朵鮮花．無論多麼努力，約翰一次只能送一隻牛回棚．而運送第第i只牛事實上需要2Ti分鐘，因爲來回都需要時間．    寫一個程序來決定約翰運送奶牛的順序，使最終被吞食的花朵數量最小．

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行輸入N，之後N行每行輸入兩個整數Ti和Di．

Output

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

    一個整數，表示最小數量的花朵被吞食．

Sample Input

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

Sample Output

86

HINT

    約翰用6，2，3，4，1，5的順序來運送他的奶牛．

Source

Silver

題解：

  簡單的貪心，最好的方法一定是先運送距離短，喫的多的奶牛。於是可以按照奶牛的飯量和運送時間的比值進行排序了，之後再用前綴和之類的求一下就可以了。

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int MAXN=100001;
struct xx
{
    int d,t;
    double s;
}a[MAXN];
bool cmp(xx a,xx b)
{
    return a.s>b.s;
}
long long tot,ans;
int main(int argc, char *argv[])
{
    int n,m,i,j;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    scanf("%d%d",&a[i].t,&a[i].d),a[i].s=a[i].d*1.0/a[i].t*1.0,tot+=a[i].d;
    sort(a+1,a+n+1,cmp);
    for(i=1;i<=n;i++) tot-=a[i].d,ans+=a[i].t*2*tot;
    printf("%lld\n",ans);
    return 0;
}