今天本来解决的很好，本来可以不聊那么结束，但是我想更完美一点，多聊几句，谁知道就聊了很长时间，很傻逼。耽误了时间！

/*************************************************************************************************/

Go Home

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 400    Accepted Submission(s): 169

Problem Description

There comes the holiday, Partychen set foot on the way home. He takes some ECNU coins to hire bodyguards to prevent from being robbed before he went home. But the bodyguard takes one coin for every kilometer. If Partychen walks without bodyguard , he will be robbed one ECNU coin by every robber on every kilometer . Of course , he can choose where to hire bodyguard or where to be robbed as he like.
For example , there
are two roads on his way home and he wants to use 8 ECNU coins to hire
bodyguard , the first road takes 4 kilometers with 5 robbers ( per
kilometer ) and the second takes 5 kilometers with 6 robbers. He could
choose the last 3 kilometers on the first road and the whole kilometers
on the second road to hire bodyguard to protect him, and leave the first
kilometer on the first road to be robbed by 5 robbers, which he will be
robbed 5 ECNU coins.
Now , Partychen want to know how many ECNU coins will be robbed at least.

Input

It consists of multi-case .
Every
case starts with two integers N and M ( 0≦N≦10,000, 0≦M≦1,000,000,000 )
which means that there are N roads and M ECNU coins to hire bodyguard.
The
followed N lines contains two integers D and P (1<=D<=10,000 ,
0<=P<=10 ) , which means the length of every road and the number
of robbers in every kilometer on this road.
End with N=0 and M=0 .

Output

An integer means the number of ECNU coins to be robbed at least.

Sample Input

2 8
4 5
5 6
3 1
5 10
5 10
5 10
0 0

Sample Output

5
140

Source

2009 Multi-University Training Contest 18 - Host by ECNU

Recommend

lcy

贪心，暗每千米的强盗数降序排即可

#include
#include
#include
#include
#include
#include
using namespace std;
#define N 12345
struct node
{
int a,b;
}c[N];
int cmp(node n1,node n2)
{
return n1.b > n2.b;
}
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m)&&(n+m))
{
int sum=;
for(int i=;i=c[i].a)
{
sum-=c[i].b*c[i].a;
m-=c[i].a;
}
else
{
sum-=c[i].b*m;
break;
}
}
cout<<sum<<endl;
}
return ;
}