[BZOJ1634]&&[Usaco2007 Jan]Protecting the Flowers 护花

描述 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.

约翰留下他的 N 只奶牛上山采木．他离开的时候，她们像往常一样悠闲地在草场里吃草．可是，当他回来的时候，他看到了一幕惨剧：牛们正躲在他的花园里，啃食着他心爱的美丽花朵！为了使接下来花朵的损失最小，约翰赶紧采取行动，把牛们送回牛棚． 牛们从 1 到 N 编号．第 i 只牛所在的位置距离牛棚 Ti(1≤Ti《2000000) 分钟的路程，而在约翰开始送她回牛棚之前，她每分钟会啃食 Di(1≤Di≤100) 朵鲜花．无论多么努力，约翰一次只能送一只牛回棚．而运送第第 i 只牛事实上需要 2Ti 分钟，因为来回都需要时间． 写一个程序来决定约翰运送奶牛的顺序，使最终被吞食的花朵数量最小．

输入格式 InputFormat

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．

输出格式 OutputFormat

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．

样例输入 SampleInput

13
82 77
97 49
73 77
60 100
94 24
31 74
5 76
24 41
100 70
97 89
38 68
41 93
89 16

样例输出 SampleOutput

362670

来源 Source

Silver

BZOJ 1634

代码 Code

贪心策略，当 t[a]/d[b]>t[a]/t[b] 时，b 放在 a 前面更优。。另外 getchar() 果然就是快。。。

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define inf 0x7fffffff
struct cow
{
    int t,d;
};
cow a[100005];
int i,j,t,n,m,l,r,z,y,x;
long long ans,tot;
inline int read()
{
    int s=0;
    char ch=getchar();
    while (ch<'0' || ch>'9') ch=getchar();
    while (ch>='0' && ch<='9') s=s*10+ch-'0',ch=getchar();
    return s;
}
inline bool comp(cow a,cow b)
{
    return a.t*b.d<a.d*b.t;
}
int main()
{
    n=read();
    tot=0;
    for (i=1;i<=n;i++) 
    {
        a[i].t=read();a[i].d=read();
        tot+=a[i].d;
    }
    sort(a+1,a+n+1,comp);
    ans=0;
    for (i=1;i<=n;i++)
    {
        tot-=a[i].d;
        ans+=(long long)2*a[i].t*tot;
    }
    printf("%lld\n",ans);
    return 0;
}