FatMouse's Speed

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16063    Accepted Submission(s): 7079
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

 

/*HDU 1160题意：n只老鼠，给出每只老鼠的weight和speed，求一排weight严格增加，speed严格递减的最长老鼠串。根据weight排序一下，然后就是一道最长不上升子序列的题。*/#include<stdio.h>#include<algorithm>using namespace std;#define MAXN 1000struct Node{    int w,s;//重量和速度     int index;//最初的序号，避免排序后乱掉顺序，后面需要输出的 }mouse[MAXN+10];bool cmp(Node a,Node b)//先按照w从小到大排序，再按照y从大到小排序 {    if(a.w<b.w)  return 1;    else if(a.w==b.w&&a.s>b.s)return 1;    else return 0;}    int dp[MAXN+10];//dp[i]表示以第i个数据结尾的符合要求的子列长度int pre[MAXN+10];//记录i对应的上一个数据int res[MAXN+10];//存放最终结果下标 int main(){    //freopen("test.in","r",stdin);    //freopen("test.out","w",stdout);    int i=1,j;    while(scanf("%d%d",&mouse[i].w,&mouse[i].s)!=EOF)    {        dp[i]=1;        pre[i]=0;        mouse[i].index=i;        i++;    }        int n=i-1;    sort(mouse+1,mouse+1+n,cmp);    int maxlen=0;//最长序列长度     int maxi;//最长序列的最后一个数下标     dp[1]=1;    for(i=1;i<=n;i++)    {        for(j=1;j<i;j++)            if(mouse[i].w>mouse[j].w&&mouse[i].s<mouse[j].s&&dp[j]+1>dp[i])            {                 dp[i]=dp[j]+1;                 pre[i]=j;                 if(dp[i]>maxlen)                 {                       maxi=i;                       maxlen=dp[i];                                       }                }                     }    int t=maxi;    i=0;    while(t!=0)    {        res[i++]=t;        t=pre[t];    }     printf("%d\n",i);    while(i>0)    {        i--;        printf("%d\n",mouse[res[i]].index);    }      return 0;             }