思维 hdu 5904 （LCIS）

LCIS

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

 

题意：给出两个序列，求最长公共递增序列。

题解：关键是递增，所以dp[ i ] =max ( dp[ i-1 ]+1,dp[ i ] ).

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int n,t,m,a[101010],b[101010],dp1[101010],dp2[101010];int main(){scanf ("%d",&t);while (t--){int i;scanf ("%d %d",&n,&m);memset(dp1,0,sizeof(dp1));memset(dp2,0,sizeof(dp2));for (i=1; i<=n; i++)scanf ("%d",&a[i]);for (i=1; i<=m; i++)scanf ("%d",&b[i]);for (i=1;i<=n;i++){dp1[a[i]]=max(dp1[a[i]],dp1[a[i]-1]+1);}for (i=1;i<=m; i++){dp2[b[i]]=max(dp2[b[i]],dp2[b[i]-1]+1);}int ans=0;for (i=1; i<=n; i++){ans=max(ans,min(dp1[a[i]],dp2[a[i]]));}printf ("%d\n",ans);}return 0;}