HDU 1160 ZOJ 1108 FatMouse's Speed

题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=1160

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
      
      
      
      

      
      
      
      解题思
      
      
      
      路：
      
      
      
      

      
      
      
      首先根据weight的数值由小到大排序，然后再以speed为标准，求最长下降子序列，思路还是DP，但是要标记子序列的下标
      
      
      
      

      
      
      
      麻烦多了，我的做
      
      
      
      法：
      
      
      
      

      
      
      
      定义MaxInclud
      
      
      
      eEnd数组，令MaxIncludeEnd[i]表示以a[i]结尾的最长下降子序列的长度，那么
      
      
      
      

      
      
      
      MaxIncludeEnd[i]=max{ MaxIncludeEnd[i] , a[j]>a[i] ? (MaxIncludeEnd[j]+1) ：-1; }
      
      
      
      

      
      
      
      其中j=0:i-1;MaxIncludeEnd[i]的初始值为1,每一个i结束后用max和MaxIncludeEnd[i]比较，更新max和下标max_flg。
      
      
      
      

      
      
      
      因为要标记下标，在计算MaxIncludeEnd[i]后，用flg[i]表示以a[i]结尾的最长下降子序列的前一个元素的下标，这样计算完
      
      
      
      

      
      
      
      MaxIncludeEnd[n]并更新完max和max_flg以后，循环递推一次就可以得到所有的元素下标，即flg[max_flg],flg[ flg[max_flg] ],……
      
      
      
      

      
      
      
      当然这里的下标是逆向的，要设置个新的数组来将其逆转。
      
      
      
      

      
      
      
      ps：如果从后往前计算最长上升子序列
      
      
      
      ，那就不
      
      
      
      必如此麻烦，
      
      
      
      减少空间。 
       
 
         
         
        code 
        
 #include<cstdio>
#include<algorithm>
using namespace std;
struct mouse
{
    int weight;
    int speed;
    int num;
}mice[1002];
int MaxIncludeEnd[1002],flg[1002],rst[1002];
int cmp(struct mouse a,struct mouse b)
{
    if(a.weight<b.weight)return 1;
    return 0;
}
int main()
{
    int w,s,i=0,j,k,size,max,max_flg;
    while(scanf("%d%d",&w,&s)!=EOF)
    {
        mice[i].weight=w;
        mice[i].speed=s;
        mice[i].num=i+1;
        i++;
    }
    size=i;
    sort(mice,mice+size,cmp);
    max=max_flg=flg[0]=0;
    for(i=0;i<size;i++)
    {
        MaxIncludeEnd[i]=1;
        for(j=0;j<i;j++)
        {
            if(mice[i].speed<mice[j].speed && mice[i].weight>mice[j].weight && MaxIncludeEnd[j]+1>MaxIncludeEnd[i])
            {
                MaxIncludeEnd[i]=MaxIncludeEnd[j]+1;
                flg[i]=j;
            }
        }
        if(max<MaxIncludeEnd[i])
        {
            max=MaxIncludeEnd[i];
            max_flg=i;
        }
    }
    printf("%d\n",max);
    k=0;
    while(max--)
    {
        rst[k++]=max_flg;
        max_flg=flg[max_flg];
    }
    for(k--;k>=0;k--)
        printf("%d\n",mice[rst[k]].num);
    return 0;
}