HDU-FatMouse's Speed

FatMouse's Speed

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15490    Accepted Submission(s): 6830
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

 

解题思路

此题要进行两个步骤，一，最长递增子序列，二，如何按顺序输出选中的最长公共子序列

 

Recommend

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
    int weight;
    int speed;
    int no;  //保存原始序号
}ch[1005];
struct node1
{
    int num;
    int pre;  //记录当前序列的前一个序列的下标，相当于指针
}dp[1005];
bool cmp(node a,node b)
{
    if(a.weight==b.weight)
        return a.speed>b.speed;
    return a.weight<b.weight;
}


int main()
{
    int n=1;
    while(scanf("%d%d",&ch[n].weight,&ch[n].speed)!=EOF)
    {
        ch[n].no=n;
        n++;
    }
    n--;
    sort(ch+1,ch+1+n,cmp);//按体重增大进行排列，若体重相同，速度从大到小排列
    memset(dp,0,sizeof(dp));
    for(int i=1;i<=n;i++)
        dp[i].num=1;  //最低选中一组
    int max=0;

    int t=1;

//此处运用长递增子序列的算法

    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<i;j++)
        {
            if(ch[j].speed>ch[i].speed&&ch[j].weight<ch[i].weight)
                {
                    if(dp[i].num<dp[j].num+1)
                    {
                        dp[i].num=dp[j].num+1;
                        dp[i].pre=j;  //记录前一组的下标
                    }
                }
        }
        if(max<dp[i].num)
        {
            max=dp[i].num;
            t=i;//找出最后的下标
        }


    }
    cout<<max<<endl;
    int  m[1005];
    for(int i=1;i<=max;i++)
    {
        m[i]=t;
        t=dp[t].pre;
    }
    for(int i=max;i>=1;i--)
        printf("%d\n",ch[m[i]].no);
}