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

  

   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 

 

Recommend

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;
}