HDU-5904 LCIS(dp)

LCIS

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 329    Accepted Submission(s): 144

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

33 31 2 33 2 110 51 23 2 32 4 3 4 5 6 11 2 3 4 51 121

 

Sample Output

150

  建立两个数组，dp1[x] 代表以x结尾的最大连续长度，dp2[x] 代表以x结尾的最大连续长度，两者的最小值就表示了当前最长公共递增子序列长度，取最大值。

//数组上限大了会超时

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;const int N = 200005;int a[N],b[N];int main(){    int t,n,m;    cin >> t;    while(t--)    {        for(int i = 0;i < N;i++)            a[i] = b[i] = 0;        scanf("%d%d",&n,&m);        int x,y;        for(int i = 0;i < n;i++)        {            scanf("%d",&x);            a[x] = a[x-1] + 1;        }        int ans = 0;        for(int i = 0;i < m;i++)        {            scanf("%d",&y);            b[y] = b[y-1] + 1;            ans = max(ans , min(b[y] , a[y]));        }        printf("%d\n",ans);    }    return 0;}