hdu 5087 Revenge of LIS II lcs變形

點擊打開鏈接鏈接 

Revenge of LIS II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1028    Accepted Submission(s): 334

Problem Description

In computer science, the longest increasing subsequence problem is to find a subsequence of a given sequence in which the subsequence's elements are in sorted order, lowest to highest, and in which the subsequence is as long as possible. This subsequence is not necessarily contiguous, or unique.
---Wikipedia

Today, LIS takes revenge on you, again. You mission is not calculating the length of longest increasing subsequence, but the length of the second longest increasing subsequence.
Two subsequence is different if and only they have different length, or have at least one different element index in the same place. And second longest increasing subsequence of sequence S indicates the second largest one while sorting all the increasing subsequences of S by its length.

 

Input

The first line contains a single integer T, indicating the number of test cases. 

Each test case begins with an integer N, indicating the length of the sequence. Then N integer Ai follows, indicating the sequence.

[Technical Specification]
1. 1 <= T <= 100
2. 2 <= N <= 1000
3. 1 <= Ai <= 1 000 000 000

 

Output

For each test case, output the length of the second longest increasing subsequence.

 

Sample Input

3 2 1 1 4 1 2 3 4 5 1 1 2 2 2

 

Sample Output

1 3 2

Hint

For the first sequence, there are two increasing subsequence: [1], [1]. So the length of the second longest increasing subsequence is also 1, same with the length of LIS.

 

求出第二長的lcs 

普通lcs再sort是不對的

反例 ：

1 1 2

方法一：

統計到每個長度的個數

#include<cstdio>
#include<cstring>
int max(int a,int b)
{
    return a>b?a:b;
}
long long cnt[1111];
int a[1111],dp[1111];
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)
            cnt[i]=1;
        int ans=0;
        long long num=0;
        for(int i=1;i<=n;i++)
        {
            cnt[i]=1;
            dp[i]=1;
            for(int j=1;j<i;j++)
            {
                if(a[j]<a[i])
                {
                    if(dp[i]<dp[j]+1)
                    {
                        dp[i]=dp[j]+1;
                        cnt[i]=cnt[j];
                    }
                    else if(dp[i]==dp[j]+1)
                        cnt[i]+=cnt[j];
                }
            }
            ans=max(ans,dp[i]);
        }
        for(int i=1;i<=n;i++)
        {
            if(dp[i]==ans)
                num+=cnt[i];
        }
        if(num==1)
            printf("%d\n",ans-1);
        else if(num>1)
            printf("%d\n",ans);
    }
    return 0;
}

方法二：

求lcs的第K優解 開二維的dp

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[1111][3];
int a[1111];
int ans[2222];
int main()
{
    int t,n;
    int s[10];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        memset(dp,0,sizeof(dp));
        int xx=0;
        for(int i=1;i<=n;i++)
        {
            dp[i][0]=1;
            dp[i][1]=0;
            for(int j=1;j<i;j++)
            {
                if(a[j]<a[i])
                {
                    s[0]=dp[j][0]+1;
                    s[1]=dp[j][1]+1;
                    s[2]=dp[i][0];
                    s[3]=dp[i][1];
                    sort(s,s+4);
                    dp[i][0]=s[3];
                    dp[i][1]=s[2];
                }
            }
            ans[xx++]=dp[i][0];
            ans[xx++]=dp[i][1];
        }
        sort(ans,ans+2*n);
        printf("%d\n",ans[xx-2]);
    }
    return 0;
}