HDU 6078 dp

题目

Wavel Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 644    Accepted Submission(s): 335

Problem Description

Have you ever seen the wave? It's a wonderful view of nature. Little Q is attracted to such wonderful thing, he even likes everything that looks like wave. Formally, he defines a sequence a1,a2,...,an as ''wavel'' if and only if a1<a2>a3<a4>a5<a6...



Picture from Wikimedia Commons


Now given two sequences a1,a2,...,an and b1,b2,...,bm, Little Q wants to find two sequences f1,f2,...,fk(1≤fi≤n,fi<fi+1) and g1,g2,...,gk(1≤gi≤m,gi<gi+1), where afi=bgi always holds and sequence af1,af2,...,afk is ''wavel''.

Moreover, Little Q is wondering how many such two sequences f and g he can find. Please write a program to help him figure out the answer.

 

Input

The first line of the input contains an integer T(1≤T≤15), denoting the number of test cases.

In each test case, there are 2 integers n,m(1≤n,m≤2000) in the first line, denoting the length of a and b.

In the next line, there are n integers a1,a2,...,an(1≤ai≤2000), denoting the sequence a.

Then in the next line, there are m integers b1,b2,...,bm(1≤bi≤2000), denoting the sequence b.

 

Output

For each test case, print a single line containing an integer, denoting the answer. Since the answer may be very large, please print the answer modulo 998244353.

 

Sample Input

1
3 5
1 5 3
4 1 1 5 3

 

Sample Output

10
Hint
(1)f=(1),g=(2).
(2)f=(1),g=(3).
(3)f=(2),g=(4).
(4)f=(3),g=(5).
(5)f=(1,2),g=(2,4).
(6)f=(1,2),g=(3,4).
(7)f=(1,3),g=(2,5).
(8)f=(1,3),g=(3,5).
(9)f=(1,2,3),g=(2,4,5).
(10)f=(1,2,3),g=(3,4,5).

题目大意

 看hint应该很好理解

解题思路

有点求最长公共子序列的意思，通过DP求解该点之前的波峰和波谷的数量。有机会再写一次吧

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define LL long long
#define read(a) scanf("%d",&a)
const int mod=998244353;
int a[2000+10];
int b[2000+10];
LL dp[2000+10][2];
LL sum[2000+10][2];
void work()
{
    memset(dp,0,sizeof(dp));
    memset(sum,0,sizeof(sum));
    int n,m;
    read(n),read(m);
    LL ans=0;
    for(int i=1;i<=n;i++)
        read(a[i]);
    for(int i=1;i<=m;i++)
        read(b[i]);
    for(int i=1;i<=n;i++)
    {
        LL peak=1;
        LL val=0;
        for(int j=1;j<=m;j++)
        {
            if(a[i]==b[j])
            {
                dp[j][0]=peak;
                dp[j][1]=val;
                ans=(ans+(peak+val)%mod)%mod;
                sum[j][0]=(sum[j][0]+dp[j][0])%mod;
                sum[j][1]=(sum[j][1]+dp[j][1])%mod;
            }
            else if(a[i]>b[j])
            {
                val=(val+sum[j][0])%mod;
            }
            else
            {
                peak=(peak+sum[j][1])%mod;
            }

        }

    }
    printf("%lld\n",ans);
}
int main()
{
    int T;
    read(T);
    while(T--)
    {
        work();
    }

    return 0;
}