LCIS 最长连续公共子序列问题-优快云博客

本文介绍了一个经典的编程问题——最长连续公共子序列（LCIS）问题，并提供了一种通过一次扫描来解决该问题的有效方法。该方法适用于两个正整数序列，能够找出它们之间的最长连续公共子序列的长度。

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

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：

给出两个长度为n、m的正整数序列，求最长连续公共子序列

思路：

简单扫一扫

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<list>
#include<stack>
#include<iomanip>
#include<cmath>
#include<bitset>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef long double ld;
#define INF (1ll<<60)-1
#define Max 1e9
using namespace std;
int T;
int n,m;
int a[100100],b[100100];
int vis[1000100],num[1000100];
int vis2[1000100],num2[1000100];
int main(){
    scanf("%d",&T);
    int mx=0;
    while(T--){
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++) scanf("%d",&a[i]),mx=max(mx,a[i]);
        for(int i=1;i<=m;i++) scanf("%d",&b[i]),mx=max(mx,b[i]);
        for(int i=1;i<=n;i++){
            if(vis[a[i]-1]) {
                num[a[i]]=num[a[i]-1]+1;

            } else {
                num[a[i]]=1;
            }
            vis[a[i]]=1;
        }
        for(int i=1;i<=m;i++){
            if(vis2[b[i]-1]) {
                num2[b[i]]=num2[b[i]-1]+1;

            } else {
                num2[b[i]]=1;
            }
            vis2[b[i]]=1;
        }
        int ans=0;
        for(int i=1;i<=1000000;i++){
            ans=max(ans,min(num[i],num2[i]));
        }
        printf("%d\n",ans);
        for(int i=1;i<=n;i++) {
            vis[a[i]]=num[a[i]]=0;
        }
        for(int i=1;i<=m;i++) {
            vis2[b[i]]=num2[b[i]]=0;
        }
    }
    return 0;
}
/*
1
4
6 10 15 30

*/