hdu6078 Wavel Sequence（分析优化递推过程，好题）

题意：

序列a有n个数，序列b有m个数。问，两个序列有多少个公共子序列满足“波浪状”。“波浪状”数组形如 a1 < a2 > a3 < a4 > a5 ……

1 <= n,m <= 2000

分析：

看了一个博客分析的非常好
http://blog.csdn.net/weyoungg/article/details/76735843

代码就是按照这个博主的思路，只不过用滚动数组把第一维滚掉了。

代码：

#include <iostream>#include <algorithm>#include <queue>#include <stack>#include <vector>#include <set>#include <cmath>#include <cstdlib>#include <cstring>#include <cstdio>using namespace std;#define ms(a,b) memset(a,b,sizeof(a))typedef long long ll;const int MAXN = 2005;const double EPS = 1e-8;const int INF = 0x3f3f3f3f;const int MOD = 998244353;int n, m, a[MAXN], b[MAXN];ll f[MAXN][2];int main() {    int T;    scanf("%d", &T);    while (T--) {        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]);        ms(f, 0);        ll ans = 0;        //f[][1] 波峰 f[][0] 波谷        for (int i = 1; i <= n; i++) {            ll top = 1, btm = 0;            for (int j = 1; j <= m; j++) {                if (a[i] == b[j]) {                    f[j][0] += top;                    f[j][1] += btm;                    ans = (ans + top + btm) % MOD;                } else if (a[i] > b[j]) {                    (btm += f[j][0]) %= MOD;                } else {                    (top += f[j][1]) %= MOD;                }            }        }        printf("%lld\n",ans);    }    return 0;}