题目链接：

LCIS

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 458 Accepted Submission(s): 212

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

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

Sample Output

题意：

求这两个序列的最长公共子序列且是一段连续的值得最长长度;

思路：

两个序列A,B，fa[i]表示A序列中以i为结尾的最长的子序列的最长长度,fb[i]也是一样,然后取max(min(fa[i],fb[i]));

AC代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <bits/stdc++.h>

#include <stack>

#include <map>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)

#define mst(ss,b) memset(ss,b,sizeof(ss));

#define lson o<<1

#define rson o<<1|1

typedef  long long LL;

template<class T> void read(T&num) {

    char CH; bool F=false;

    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());

    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());

    F && (num=-num);

}

int stk[70], tp;

template<class T> inline void print(T p) {

    if(!p) { puts("0"); return; }

    while(p) stk[++ tp] = p%10, p/=10;

    while(tp) putchar(stk[tp--] + '0');

    putchar('\n');

}

const LL mod=1e9+7;

const double PI=acos(-1.0);

const LL inf=1e18;

const int N=1e6+10;

const int maxn=1e5+10;

const double eps=1e-12;

int n,m,a[maxn],b[maxn],fa[N],fb[N];

int main()

{

    int t;

    read(t);

    while(t--)

    {

        read(n);read(m);

        For(i,1,n)read(a[i]);

        For(i,1,m)read(b[i]);

        For(i,1,n)fa[a[i]-1]=fa[a[i]]=0,fb[a[i]-1]=fb[a[i]]=0;

        For(i,1,m)fa[b[i]-1]=fa[b[i]]=0,fb[b[i]-1]=fb[b[i]]=0;

        For(i,1,n)fa[a[i]]=fa[a[i]-1]+1;

        For(i,1,m)fb[b[i]]=fb[b[i]-1]+1;

        int ans=0;

        for(int i=1;i<=n;i++)ans=max(ans,min(fa[a[i]],fb[a[i]]));

        printf("%d\n",ans);

    }

    return 0;

}

巴特西

hdu-5904 LCIS(水题)

LCIS

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