FatMouse's Speed（dp）

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

 

 

Source

Zhejiang University Training Contest 2001

 

Recommend

Ignatius

 

此题目为经典的动态规划题目，让你寻找一组最长的随着体重增加而速度减小的，输出它们的编号！

代码;

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#include <iomanip>#define MAXN 1000#define maxn 15#define mod 1000000007#define INF 0x3f3f3f3f#define exp 1e-6#define pi acos(-1.0)using namespace std;struct Node    //建立一个结构体，存放数据{    int w,s;    int index;}mouse[MAXN+10];bool cmp(Node a,Node b)   //自定义排序，先按体重有小到大在速度大到小{    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];   //存放最终结果的下标int main(){    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];  //最大                 }            }    }    int t=maxi;    i=0;    while(t!=0)    {        res[i++]=t;        t=pre[t];   //直到前一个为0即不存在前一个了    }    printf("%d\n",i);    while(i>0)    {        i--;        printf("%d\n",mouse[res[i]].index);    }    return 0;}