HDU6078 Wavel Sequence(动态规划)

Wavel Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 728    Accepted Submission(s): 377

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

13 51 5 34 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 

 

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liuyiding

 

题意：给出两个数列，找到两个数列中成波浪状的公共子数列的个数

首先可以找到一个很明显的递推式，dp[i][j][0]表示a[i]和b[j]结尾且最后一个为波谷，dp[i][j][1]表示a[i]和b[j]结尾而且最后一个为波峰

如果a[i]!=b[j]   dp[i][j][0]=0且dp[i][j][1]等于0

否则，dp[i][j][0]等于所有p<i和q<j的dp[p][q][1]的情况加起来，dp[i][j][1]亦然

所以我们用两个数组记录，sum[i][j][0]表示以b[j]结尾，所有k<=i的dp[k][j][0]的情况加起来，这样就可以优化到n^2的复杂度了。

#include<bits/stdc++.h>#define fi first#define se second#define pb push_back#define CLR(A, X) memset(A, X, sizeof(A))using namespace std;typedef long long LL;typedef pair<int, int> PII;const double eps = 1e-10;int dcmp(double x){if(fabs(x)<eps) return 0; return x<0?-1:1;}const LL INF = 0x3f3f3f3f;const LL MOD = 998244353;const int N = 2e3+5;int b[N], a[N];LL sum[N][2], dp[N][2];int main() {//    freopen("1001.txt", "r", stdin);    int T, n, m;    scanf("%d", &T);    while(T--) {        CLR(sum, 0); CLR(dp, 0);        scanf("%d%d", &n, &m);        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);        for(int i = 1; i <= m; i++) scanf("%d", &b[i]);        LL ans = 0;        for(int i = 1; i <= n; i++) {            LL cnt1 = 1, cnt0 = 0;            for(int j = 1; j <= m; j++) {                dp[j][0] = dp[j][1] = 0;                if(a[i] == b[j]) {                    dp[j][0] = cnt1;                    dp[j][1] = cnt0;                    ans = (ans+cnt1+cnt0)%MOD;                }                else if(b[j] < a[i]) (cnt0 += sum[j][0]) %= MOD;                else (cnt1 += sum[j][1]) %= MOD;            }            for(int j = 1; j <= m; j++) {                if(b[j] == a[i]) {                    (sum[j][0] += dp[j][0]) %= MOD;                    (sum[j][1] += dp[j][1]) %= MOD;                }            }        }        printf("%lld\n", ans);    }    return 0;}