2017多校训练Contest4: 1012 Wavel Sequence hdu6078

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概述

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

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).

设 $f_{i,j,k}$ 表示仅考虑 $a [1 . . i]$ 与 $b [1 . . j]$ ，选择的两个子序列结尾分别是 $a_i$ 和 $b_j$ ，且上升下降状态是 $k$ 时的方案数，则 $f_{i,j,k}=sum f_{x,y,1-k}$ ，其中 $x < i, y < j$ 。暴力转移的时间复杂度为 $O(n^4)$ ，不能接受。

考虑将枚举决策点 $x, y$ 的过程也DP掉。设 $g_{i,y,k}$ 表示从某个 $f_{x,y,k}$ 作为决策点出发，当前要更新的是 $i$ 的方案数， $h_{i,j,k}$ 表示从某个 $f_{x,y,k}$ 作为决策点出发，已经经历了 $g$ 的枚举，当前要更新的是 $j$ 的方案数。转移则是要么开始更新，要么将 $i$ 或者 $j$ 继续枚举到 $i + 1$ 以及 $j + 1$ 。因为每次只有一个变量在动，因此另一个变量恰好可以表示上一个位置的值，可以很方便地判断是否满足上升和下降。

时间复杂度 $O(n^2)$ 。

以上是官方题解

然后我们的程序复杂度似乎不是n^2的。中间用二维树状数组维护转移所以多了两个log

#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cassert>
using namespace std;
const int N=2110;
const int M=2005;
const int pp=998244353;
int tre[2][N][N];
int f[2][N][N];
int a[N],b[N];
int n,m;
void add(int &x,int y)
{
x=x+y;
if (x>pp) x-=pp;
}
void work(int now,int xx,int yy,int v)
{
for (int x=xx;x<=M;x+=x&-x)
for (int y=yy;y<=M;y+=y&-y)
add(tre[now][x][y],v);
}
int get(int now,int xx,int yy)
{
int s=0;
for (int x=xx;x;x-=x&-x)
for (int y=yy;y;y-=y&-y)
add(s,tre[now][x][y]);
return s;
}
int main()
{
int T;
scanf("%d",&T);
while (T--)
{
scanf("%d%d",&n,&m);
for (int i=0;i<=M;i++)
for (int j=0;j<=M;j++) tre[0][i][j]=tre[1][i][j]=0;
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++)
f[0][i][j]=f[1][i][j]=0;
for (int i=1;i<=n;i++)
scanf("%d",&a[i]);
for (int i=1;i<=m;i++)
scanf("%d",&b[i]);
int an=0;
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++)
if (a[i]==b[j])
{
add(f[0][i][j],(get(1,j-1,M)-get(1,j-1,a[i])+pp)%pp);
add(f[1][i][j],get(0,j-1,a[i]-1));
add(f[0][i][j],1);
work(0,j,a[i],f[0][i][j]);
work(1,j,a[i],f[1][i][j]);
add(an,f[0][i][j]);
add(an,f[1][i][j]);
}
printf("%dn",(an+pp)%pp);
}
}