Revenge of LIS II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1427 Accepted Submission(s): 489

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.

Source

BestCoder Round #16

题意:求解一个串中次长上升子序列的的长度。

题解:主要是求出有多少个最长上升子序列,如果个数大于1,那么次长就是最长,如果为1 ，那么次长就为最长的减一。关键是怎么统计最长上升子序列的个数？？这里用到一个数组记录dp[i]的个数,当dp[j]+1>dp[i]时,dp[i]的个数为num[j],如果dp[j]+1==dp[i]，那么 num[i] = num[i]+num[j]

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<cmath>

using namespace std;

const int N = ;

int dp[N],num[N];

int a[N];

int main(){

    int tcase;

    scanf("%d",&tcase);

    while(tcase--){

        int n;

        scanf("%d",&n);

        for(int i=;i<=n;i++){

            scanf("%d",&a[i]);

        }

        int ans = dp[] = num[]= ;

        for(int i=;i<=n;i++){

            dp[i] =num[i]= ;

            for(int j=;j<i;j++){

                if(a[i]>a[j]){

                    if(dp[i]==dp[j]+){

                        num[i]+=num[j];

                   }

                   if(dp[i]<dp[j]+){

                        dp[i] = dp[j]+;

                        num[i] = num[j];

                   }

                }

            }

            ans = max(ans,dp[i]);

        }

        int cnt=;

        for(int i=;i<=n;i++){

            if(dp[i]==ans){

                cnt+=num[i];

            }

        }

        if(cnt==) printf("%d\n",ans-);

        else printf("%d\n",ans);

    }

    return ;

}

巴特西

hdu 5087(次长上升子序列)

Revenge of LIS II

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