HDU 6078 Wavel Sequence 树状数组优化DP

Wavel Sequence

 

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).

 

 

题解：

　　设定dp[i][j][0/1] 表示已a[i], a[j]结尾的 序列，长度为奇数1，偶数0的方案数

　　假设 当前a[i] = a[j] = x；

　　那么dp[i][j][1]就要继承   满足尾端值比当前x大的 那些位置的那些dp[i'][j'][0]    (1<=i' < i && 1<=j' < j)

　　同理dp[i][j][0] 就要继承 满足结尾值比x小的那些 dp[i'][j'][1]    1<=i' < i && 1<=j' < j

　　定义树状数组sum[i][0] 表示 以i值结尾的 长度为奇数偶数的方案

　　在树状数组上面修改查询即可

代码：

　　

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
typedef unsigned long long ULL;
const long long INF = 1e18+1LL;
const double pi = acos(-1.0);
const int N = 2e5+10, M = 1e3+20,inf = 2e9;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

const LL mod = 998244353LL;

int n,m,a[N],b[N];
int nex[N],head[N],last[N],fi[N],vis[N],mx;
LL sum[N][2],dp[2005][2005][2];
void update(int x,LL c,int p) {
    for(int i = x; i <= mx; i += i&(-i))
        sum[i][p] += c,sum[i][p] %= mod;
}
LL ask(int x,int p) {
    LL ret = 0;
    if(x == 0) return 0;
    for(int i = x; i; i -= i&(-i))
        ret += sum[i][p],ret %= mod;
    return ret;
}
int main() {
    int T;
    T = read();
    while(T--) {
        n = read();
        m = read();
        mx = -1;
        for(int i = 1; i <= n; ++i) a[i] = read(),mx = max(mx,a[i]);
        for(int i = 1; i <= m; ++i) b[i] = read(),mx = max(mx,b[i]);
        for(int i = 0; i <= n; ++i)
            for(int j = 1; j <= m;++j)
                for(int k = 0; k < 2; ++k)
                    dp[i][j][k] = 0;

        LL ans = 0,tmp1,tmp2;
        for(int i = 1; i <= n; ++i)
        {
            for(int j = 1; j <= m; ++j)
                dp[i][j][0] += dp[i-1][j][0],dp[i][j][0] %= mod,
                dp[i][j][1] += dp[i-1][j][1],dp[i][j][1] %= mod;
            for(int j = 1; j <= mx; ++j) sum[j][0] = 0, sum[j][1] = 0;

            for(int j = 1; j <= m; ++j) {
                if(a[i] == b[j]) {

                    tmp1 = ask(a[i]-1,1) % mod;
                    tmp2 = (ask(mx ,0) - ask(a[i],0) + mod) % mod;

                    dp[i][j][0] += tmp1;
                    dp[i][j][0] %= mod;
                    dp[i][j][1] += (tmp2+1LL) % mod;
                    dp[i][j][1] %= mod;

                    ans += ((tmp1+tmp2)%mod+1LL) % mod;
                    ans %= mod;
                }
                if(dp[i-1][j][0])
                    update(b[j],dp[i-1][j][0],0);
                if(dp[i-1][j][1])
                    update(b[j],dp[i-1][j][1],1);
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}