hdu 1160 FatMouse's Speed

FatMouse's Speed

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6357    Accepted Submission(s): 2780
Special Judge

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. 

 

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
  
  
  
  


  
  
  
  

   
   设Mice[i].W表示第i只老鼠的重量，Mice[i].S表示第i只老鼠的速度。我们先对Mice进行排序，以W为第一关键字，从小到大，S为第二关键字，从大到小。  
设f[i]为Mice[i]至Mice[n]最长的序列长度。考虑某一个f[i]，则有： 
  
nf[i] = max(f[i], f[j]+1) (1<=j<i，且Mice[i].W> Mice[j].W，Mice[i].S < Mice[j].S) 
n其中，初始条件为f[i]=1 (i=1, 2, ..., n)。 
   
   

  
  
  
  

   
   AC代码：
  
  
  
  

   
   
   
   
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;

struct mouse                   //定义一个mouse结构体
{
     int W,S;
     int NO;
};

mouse m[10000];           //声明一个mouse类型数组
int f[10000];                    //f[]用来存到第i只老鼠时，前面最大的老鼠的数量
int g[10000];                   //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)           //编译器里面的sort()内置函数，定义自己的的排序函数
{
       if(a.W==b.W) return a.S>b.S;  //如果两只mouse的体重相等，就速度快的排在前面
       else return a.W<b.W;              //两只mouse体重小的排在前面
}

int main()
{
      int i=0,j,n,max=0,flag=0;
      while(scanf("%d%d",&m[i].W,&m[i].S)!=EOF)      //输入mouse的体重，速度和编号
     {
           m[i].NO=i+1;
           i++;

     }
     n=i;
     sort(m,m+n,cmp);                                                //进行排序

     f[0]=1;g[0]=0;max=1;                                           //数据初始化
     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;
}