hdu 4442 贪心

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http://acm.hdu.edu.cn/showproblem.php?pid=4442

5
1 2
2 3
3 4
4 5
5 6
0

1419
Hint
In the Sample Input, WANGPENG just follow the given order. He spends 1 second in the first queue, 5 seconds in the 2th queue, 27 seconds in the 3th queue, 
169 seconds in the 4th queue, and 1217 seconds in the 5th queue. So the total time is 1419s. WANGPENG has computed all possible orders in his 
120-core-parallel head, and decided that this is the optimal choice.
 

/**
hdu 4442  贪心
题目大意：某人去做体检，最初每个体检项目前需要排对x秒，如果选择排其中一个，那么其他队的时间每秒增加y秒，问怎样安排排队的先后能可以使总体的
          排队时间最短
解题思路：假设有两个队伍a和b，如先去排a，则总体等待时间为：ax+ax*by+bx,如先去排b,则总体的等待时间为bx+bx*ay+ax。我们假设a在前优，则：
          ax*by<ay*bx。由此推广到多个队伍，按照ax*by<ay*bx的标准排序后所得的顺序处理即为答案。
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=100005;
typedef long long LL;
const int mod=365*24*60*60;
int n;

struct note
{
    LL x,y;
    bool operator < (const note &other)const
    {
        return x*other.y<other.x*y;
    }
}a[maxn];

int main()
{
    while(~scanf("%d",&n))
    {
        if(n==0)break;
        for(int i=0;i<n;i++)
        {
            scanf("%I64d%I64d",&a[i].x,&a[i].y);
        }
        sort(a,a+n);
        LL maxx=0;
        for(int i=0;i<n;i++)
        {
            maxx=(maxx+a[i].x+maxx*a[i].y)%mod;
        }
        printf("%I64d\n",maxx);
    }
    return 0;
}

hdu 4442 贪心

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原文地址：http://blog.csdn.net/lvshubao1314/article/details/44917589