HDU 6078 Wavel Sequence (dp)

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...

img

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

 

题意

定义波浪序列为满足 a1< a2 > a3 < a4 ... 的序列,现给出两个数组 a 和 b ,从 a 中选出满足波浪序列的一个子序列 f , b 中选出满足波浪序列的子序列 g ,求有多少种选法满足 f = g 。

 

思路

dp[0][j] 代表以 b[j] 结尾且最后为波谷的情况数目。

dp[1][j] 代表以 b[j] 结尾且最后为波峰的情况数目。

显然,结尾为波谷的情况可以由 波峰 + 一个小的数 转移而来,而结尾为波峰的情况可以由 波谷 + 一个大的数 转移而来。

因此我们定义 ans0、ans1 分别表示在该轮中相对于 a[i] 来说 b[j] 可作为波谷与波峰的数目。

 

枚举每一个 a[i] ,并且判断其与 b[j] 的大小关系:

  • a[i] < b[j] ,则说明 b[j] 可作为一个波峰出现
  • a[i] > b[j] ,则说明 b[j] 可作为一个波谷出现
  • a[i] = b[j] ,则说明找到一个对 f = g 有贡献的值,更新答案

 

AC 代码

#include<bits/stdc++.h>
using namespace std;

typedef __int64 LL;

const int mod = 998244353;
const int maxn = 2100;

int a[maxn],b[maxn];
LL dp[2][maxn];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=0; i<n; i++)
            scanf("%d",a+i);
        for(int i=0; i<m; i++)
            scanf("%d",b+i);
        memset(dp,0,sizeof(dp));
        LL ans=0;
        for(int i=0; i<n; i++)
        {
            LL ans1=1,ans0=0;   // 该轮最后一个为波峰/波谷的数量
            for(int j=0; j<m; j++)
            {
                if(a[i]==b[j])  // 当前位有两种状态(波峰、波谷),可分别由之前的波谷、波峰转移而来
                {
                    dp[1][j]+=ans0;
                    dp[0][j]+=ans1;
                    ans=(ans+ans0+ans1)%mod;
                }
                else if(a[i]<b[j])  //  b[j] 可相对于 a[i] 做波峰
                    ans1=(ans1+dp[1][j])%mod;
                else ans0=(ans0+dp[0][j])%mod;  // 反之亦然
            }
        }
        printf("%I64d\n",ans);
    }
    return 0;
}