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hdu 5904 LCIS dp

2024-08-14 16:52:53   219人阅读

LCIS

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)


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
 

 

Source
BestCoder Round #87
#include<bits/stdc++.h>using namespace std;#define ll long long#define pi (4*atan(1.0))const int N=1e5+10,M=1e6+1010,inf=1e9+10,mod=1e9+7;const ll INF=1e18+10;int n,m;int a[N],b[N];int dpn[M],dpm[M];int main(){    int T;    scanf("%d",&T);    while(T--)    {        int ans=0;        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            dpn[a[i]]=max(dpn[a[i]],dpn[a[i]-1]+1);        }        for(int i=1;i<=m;i++)        {            scanf("%d",&b[i]);            dpm[b[i]]=max(dpm[b[i]],dpm[b[i]-1]+1);            ans=max(ans,min(dpm[b[i]],dpn[b[i]]));        }        for(int i=1;i<=n;i++)        dpn[a[i]]=0;        for(int i=1;i<=m;i++)        dpm[b[i]]=0;        printf("%d\n",ans);    }}

 

hdu 5904 LCIS dp


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