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FatMouse believes that the fatter a mouse is, the faster it runs. To disprove this, you want to take the data on a collection of mice and put as large a subset of this data as possible into a sequence so that the weights are increasing, but the speeds are decreasing.
Input
Input contains data for a bunch of mice, one mouse per line, terminated by end of file.

The data for a particular mouse will consist of a pair of integers: the first representing its size in grams and the second representing its speed in centimeters per second. Both integers are between 1 and 10000. The data in each test case will contain information for at most 1000 mice.

Two mice may have the same weight, the same speed, or even the same weight and speed.
Output
Your program should output a sequence of lines of data; the first line should contain a number n; the remaining n lines should each contain a single positive integer (each one representing a mouse). If these n integers are m[1], m[2],…, m[n] then it must be the case that

W[m[1]] < W[m[2]] < … < W[m[n]]

and

S[m[1]] > S[m[2]] > … > S[m[n]]

In order for the answer to be correct, n should be as large as possible.
All inequalities are strict: weights must be strictly increasing, and speeds must be strictly decreasing. There may be many correct outputs for a given input, your program only needs to find one.
Sample Input
6008 1300
6000 2100
500 2000
1000 4000
1100 3000
6000 2000
8000 1400
6000 1200
2000 1900
Sample Output
4
4
5
9
7
先说说这个题的题意吧有人认为老鼠越胖跑的越快，给你一些数据让你找出最长的反例。并且要输出这个反例的组数。
这里就要在开始的时候在输入数据的时候要把这个每个数据的组数要保存，先对这个数据进行保存，然后按照重量从小到大排序，如果一样，那就按照速度的反序排列，因为这个题要速度从大到小排列。
这个开一个dp数列，如果复合结果那就对这个数列+1，炳耀开另一个数列保存这个的组数最后输出这个数。

#include <iostream>
#include <algorithm>
using namespace std;
int dp[1010],p[1010],s[1010];
struct node
{
    int a,b,c;
}a[1010];
int cmp(node a,node b)
{
    if(a.a!=b.a)
    return a.a<b.a;
    else return a.b>b.b;
}
int main()
{
   int i = 1;
   while(cin>>a[i].a>>a[i].b)
   {
       a[i].c=i;
       dp[i]=1;
       i++;
   }
   int n=i-1;
   sort(a+1,a+n+1,cmp);
   int len=0;
   int xi;
   dp[1]=1;
   for(i=1;i<=n;i++)
   {
       for(int j=1;j<i;j++)
       {
            if(a[i].a>a[j].a&&a[i].b<a[j].b&&dp[i]<dp[j]+1)
            {
                dp[i]=dp[j]+1;
                p[i]=j;
                if(dp[i]>len)
                {
                     len=dp[i];
                     xi=i;
                }
            }
       }
   }
   int t=xi;
   i=0;
   while(t!=0)
   {
       s[i++]=t;
       t=p[t];
   }
   cout<<i<<endl;
   while(i>0)
   {
       i--;
       cout<<a[s[i]].c<<endl;
   }
   return 0;
}