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FatMouse's Speed

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14556    Accepted Submission(s): 6429
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 13006000 2100500 20001000 40001100 30006000 20008000 14006000 12002000 1900

 

Sample Output

44597

 
题意：输入多只老鼠数据（包括质量和速度），以EOF结束。然后输出质量严格递增且速度严格递减的最大老鼠数量，并输出各老鼠的序号。（有可能有序号排列㓟多种情况，输出满足条件的一种即可）
思路：一开始明白题意后，思路很好想，就是按照质量排下序，然后按照速度求最长不递增子序列长度，并记录序号即可。但无奈有了思路后，代码却写得不好，这主要反映了自己将算法转化成代码能力还较弱，有待提高。
写代码时主要遇到的问题是记录序列的序号。首先，在结构体中定义一个num来存储原序号，以至于排列时不乱。然后用dp[i]存储到第i个数的最长序列，用p[i]存储排序后，序列长度每增加一个的位置记录，但是没法记录序列的第一个数，因此又用ans来存储第一个数的位置。

AC代码：

#include <iostream>#include<cmath>#include<cstdio>#include<cstring>#include<algorithm>#define N 1100using namespace std;struct node{    int w,s,n;}m[N];int dp[N],p[N];bool cmp(node a,node b){    return a.w<b.w;}int main(){    //freopen("D://in.txt","r",stdin);    int k,ans;    for(k=1;~scanf("%d%d",&m[k].w,&m[k].s);k++)    {        m[k].n=k;        dp[k]=1;    }    ans=k;    dp[ans]=0;    sort(m,m+k,cmp);   /* for(int i=1;i<k;i++)        printf("%d ",m[i].w);    cout<<endl;*/    memset(p,0,sizeof(p));    for(int i=k-1;i>=0;i--)    {        for(int j=k-1;j>i;j--)        {            if(m[i].w<m[j].w&&m[i].s>m[j].s&&dp[j]+1>dp[i])            {                dp[i]=dp[j]+1;                p[i]=j;                if(dp[ans]<dp[i])                    ans=i;            }        }    }  /*  for(int i=0;i<k;i++)        printf("%d %d\t",p[i],m[p[i]].n);*/    cout<<dp[ans]<<endl;    while(ans!=0)    {        cout<<m[ans].n<<endl;        ans=p[ans];    }    return 0;}