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…
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).
Source
2017 Multi-University Training Contest - Team 4
題意：
給出a，b數列，定義波浪數列爲a1＜a2>a3＜a4>a5 ＜a6…（第一個數必須小於第二個數）。
現在找出f,g，使得
想法：
dp[0][j] 代表以 b[j] 結尾且最後爲波谷的情況數目。
dp[1][j] 代表以 b[j] 結尾且最後爲波峯的情況數目。
顯然，結尾爲波谷的情況可以由 波峯 + 一個小的數 轉移而來，而結尾爲波峯的情況可以由 波谷 + 一個大的數 轉移而來。
因此我們定義 ans0、ans1 分別表示在該輪中相對於 a[i] 來說 b[j] 可作爲波谷與波峯的數目。
枚舉每一個 a[i] ，並且判斷其與 b[j] 的大小關係：
若 a[i] < b[j] ，則說明 b[j] 可作爲一個波峯出現
若 a[i] > b[j] ，則說明 b[j] 可作爲一個波谷出現
若 a[i] = b[j] ，則說明找到一個對 f = g 有貢獻的值，更新答案
想法來自這裏 講的真的清楚
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll p=998244353;
int a[2005],b[2005];
ll dp[2][2005];
int main()
{
int t;
scanf("%d",&t);
while (t--)
{
int n,m;
ll sum=0;
memset(dp,0,sizeof(dp));
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]);
for (int i=0; i<n; i++)
{
ll ans1=1; ///統計 此數爲波峯 波浪數
ll ans0=0; ///統計 此數爲波谷 波浪數
for (int j=0; j<m; j++)
{
if (a[i]==b[j]) ///相等更新對應波峯波谷的值
{
dp[0][j]+=ans1;///當前數作爲波峯接在波谷之後形成的波浪數
dp[1][j]+=ans0;///當前數作爲波谷接在波峯之後形成的波浪數
sum=(sum+ans1+ans0)%p; ///總情況數
}
else if(a[i]> b[j]) ///b小，可作爲波谷接在波峯之後形成波浪
{
ans0+=dp[0][j];
ans0%=p;
}
else
{
ans1+=dp[1][j]; ///b大，可作爲波峯接在波谷之後形成波浪
ans1%=p;
}
}
}
printf("%lld\n",sum);
}
return 0;
}
ps：攢了一堆博客沒寫嗚嗚嗚