FatMouse's Speed

Problem Description

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. 


题目大意：本题的要求是输出一个最长的排列，要求每组数据的第一个数能从 小到大排列，第二个数据能从大到小排列，然后输出最长的排列数，并输出本排列数在起始时的位置。
万恶的题目，搞的我弄出了正确的答案都不敢交，还在做死地检查自己的代码那里错了。看了学长的代码后才知道自己的答案是对的
#include<cstdio>
#include<algorithm>
using namespace std;
struct mouse                   
{
     int W,S;
     int NO;
}m[1005];
           //声明一个mouse类型数组
int f[1005];                    //f[]用来存到第i只老鼠时，前面最大的老鼠的数量 
int g[1005];                   //g[]用来存第i只老鼠前面的一只老鼠的序号
bool cmp(mouse a,mouse b)          
{
       if(a.W==b.W) return a.S>b.S;        
       else return a.W<b.W;             
}
int main()
{
      int i=0,j,n,max=0,flag=0;
      while(scanf("%d%d",&m[i].W,&m[i].S)!=EOF)     
           m[i].NO=i+1;                                         
           i++; 
     }
     n=i;
     sort(m,m+n,cmp);                                                
     for(i=1;i<n;i++) 
     {
          f[i]=1;
          for(j=i-1;j>=0;j--)
         {
              if(m[i].W>m[j].W&&m[i].S<m[j].S&&f[j]+1>f[i])     //体重要上升，速度要下降，数量要增加
              {                                                                    
                  f[i]=1+f[j];
                  g[i]=j;
              }
         }
         if(max<f[i])              //每次最大的一次数量，并记录后到数量最大的那只老鼠的排序号
         {
              max=f[i];
              flag=i;
          }
      }
 
      for(i=0,j=max-1;i<max;i++)   //将排序号倒置过来
      {
            f[j--]=flag;
            flag=g[flag];
       }
       printf("%d\n",max);
       for(i=0;i<max;i++)
          printf("%d\n",m[f[i]].NO);    //输出老鼠的编号
       return 0;
}