HDU 5904 LCIS

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5904

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

 

 

Hint:

 

官方题解：

令fif(i)是以aia​i​​结尾的最大值, gig(i)是以bib​i​​结尾的最大值. 答案就是maxaibjminfigjmax​a​i​​=b​j​​​​{min(f(i),g(j)}. ff和gg随便dp一下就出来了.

 

代码：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm> 
using namespace std;
const int maxn = 1e6+10;
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof(a))
int num1[maxn],num2[maxn];
int dp1[maxn],dp2[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        met(dp1,0);met(dp2,0);
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&num1[i]);
            dp1[num1[i]]=max(dp1[num1[i]],dp1[num1[i]-1]+1);
        }
        for(int i=0;i<m;i++)
        {
            scanf("%d",&num2[i]);
            dp2[num2[i]]=max(dp2[num2[i]],dp2[num2[i]-1]+1);
        }
        int ans=-inf;
        for(int i =0;i<n;i++)
            ans=max(ans,min(dp1[num1[i]],dp2[num1[i]]));
        printf("%d
",ans);
    }
}