HDU 5904:LCIS (LCIS)

LCIS

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 440    Accepted Submission(s): 201


Problem Description
Alex has two sequences a1,a2,...,an and b1,b2,...,bm. He wants find a longest common subsequence that consists of consecutive values in increasing order.

 

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:The first line contains two integers n and m (1≤n,m≤100000) -- the length of two sequences. The second line contains n integers: a1,a2,...,an (1≤ai≤106). The third line contains n integers: b1,b2,...,bm (1≤bi≤106).

There are at most 1000 test cases and the sum of n and m does not exceed 2×106.

 

Output
For each test case, output the length of longest common subsequence that consists of consecutive values in increasing order.

 

Sample Input
3
3 3
1 2 3
3 2 1
10 5
1 23 2 32 4 3 4 5 6 1
1 2 3 4 5
1 1
2
1

 

Sample Output
1
5
0

 

题意:给定两个序列,求它们的最长公共递增子序列的长度, 并且这个子序列的值是连续的。

分析:我们可以先处理a,b的每个数的最长连续长度,最终取两者最小值的最大值,状态转移方程:dp[a[i]] = max(dp[a[i]], dp[a[i]-1] + 1);

 

AC代码:

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<iostream>
using namespace std;
#define MMAX 110000
int a[MMAX],b[MMAX],dpa[MMAX],dpb[MMAX];
int main()
{
    cout.sync_with_stdio(false);
    int T;
    cin>>T;
    while(T--)
    {
        int n,m;
        cin>>n>>m;
        memset(dpa,0,sizeof(dpa));
        memset(dpb,0,sizeof(dpb));
        int maxx=0;
        for(int i=1; i<=n; i++)
        {
            cin>>a[i];
            dpa[a[i]]=dpa[a[i]-1]+1;
            maxx=max(maxx,a[i]);
        }
        for(int i=1; i<=m; i++)
        {
            cin>>b[i];
            dpb[b[i]]=dpb[b[i]-1]+1;
            maxx=max(maxx,b[i]);
        }
        int ans=0;
        for(int i=1; i<=maxx; i++)
            ans=max(ans,min(dpa[i],dpb[i]));
        cout<<ans<<endl;
    }
}