HDU 6078 Wavel Sequence（dp）

Description
定义波浪序列：a1 > a2 < a3…，现在给出一个长度为n的序列a和一个长度为m的序列b，求a和b的公共波浪子序列个数
Input
第一行一整数T表示用例组数，每组用例首先输入两个整数n和b表示两个序列的长度，然后输入一个长度为n的序列a[1]~a[n]和一个长度为m的序列b[1]~b[m] (1<=T<=15,1<=n,m,a[i],b[j]<=2000)
Output
输出a和b的公共波浪子序列个数
Sample Input
1
3 5
1 5 3
4 1 1 5 3
Sample Output
10
Solution

Code

#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=2001,mod=998244353;
namespace fastIO
{#define BUF_SIZE 100000//fread -> readbool IOerror=0;inline char nc() {static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;if(p1==pend) {p1=buf;pend=buf+fread(buf,1,BUF_SIZE,stdin);if(pend==p1) {IOerror=1;return -1;}}return *p1++;}inline bool blank(char ch) {return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';}inline void read(int &x) {char ch;while(blank(ch=nc()));if(IOerror)return;for(x=ch-'0';(ch=nc())>='0'&&ch<='9';x=x*10+ch-'0');}#undef BUF_SIZE
};
using namespace fastIO;
int T,n,m,a[maxn],b[maxn],dp[maxn][2],sum[maxn][2];
int inc(int x,int y)
{return x+y>=mod?x+y-mod:x+y;
}
int main()
{read(T);while(T--){read(n),read(m);for(int i=1;i<=n;i++)read(a[i]);for(int i=1;i<=m;i++)read(b[i]);memset(sum,0,sizeof(sum));int ans=0;for(int i=1;i<=n;i++){int num0=0,num1=1;for(int j=1;j<=m;j++){dp[j][0]=dp[j][1]=0;if(a[i]==b[j]){dp[j][0]=num1,dp[j][1]=num0;ans=inc(ans,inc(num0,num1)); }else if(a[i]<b[j])num1=inc(num1,sum[j][1]);else num0=inc(num0,sum[j][0]);}for(int j=1;j<=m;j++)sum[j][0]=inc(sum[j][0],dp[j][0]),sum[j][1]=inc(sum[j][1],dp[j][1]);}printf("%d\n",ans);}return 0;
}

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