(2017多校训练第四场)HDU - 6078 Wavel Sequence dp

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定义状态dp[i][j][0]表示以a[i]，b[j]结尾的且为波谷的情况总和，dp[i][j][1] 为波峰。

对于某个i，j满足a[i] == b[j]，则dp[i][j][0] = sum(dp[x][y][1]), x < i && y < j && b[y] > a[i]

设sum[i-1][y][1] = ∑dp[x][y][1] , x <= i-1

则dp[i][j][0] = ∑sum[i-1][y][1], b[y] > a[i]

对于每一个b[j]，sum[i][j]累计了前i个的a[i]的影响，每次求出dp[i][j]后O(1)更新即可，即sum[i][j][1] =sum[i-1][j][1] + dp[i][j][1]。

然后对于某一个a[i]的所有b[j]，可以以前缀和的形式，利用sum[i-1][y]，O(n)全部更新。

所以复杂度O(nm)。

代码如下：

#include <bits/stdc++.h>using namespace std;
typedef long long int LL;
const int MOD = 998244353;
const int N = 2005;
int a[N], n;
int b[N], m;
int dp[N][2];
int sum[N][2];int main()
{//freopen("test.txt", "r", stdin);//freopen("out.txt", "w", stdout);cin.sync_with_stdio(false);int T;cin >> T;while (T--){cin >> n >> m;for (int i = 1; i <= n; i++)cin >> a[i];for (int i = 1; i <= m; i++)cin >> b[i];memset(dp, 0, sizeof(dp));memset(sum, 0, sizeof(sum));LL ans = 0;for (int i = 1; i <= n; i++){int cnt1 = 0;int cnt0 = 0;for (int j = 1; j <= m; j++){if (a[i] == b[j]){dp[j][0] = cnt1 + 1; // 因为波谷可以作为开头，所以自身算一种。dp[j][1] = cnt0;ans = (ans + cnt0 + cnt1 + 1) % MOD;}else if (b[j] > a[i])cnt1 = (cnt1 + sum[j][1]) % MOD;elsecnt0 = (cnt0 + sum[j][0]) % MOD;}for (int j = 1; j <= m; j++)if (a[i] == b[j]){sum[j][0] = (sum[j][0] + dp[j][0]) % MOD;sum[j][1] = (sum[j][1] + dp[j][1]) % MOD;}}cout << ans << endl;}return 0;
}

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