Acwing-----272. 最长公共上升子序列

• 链接：https://www.acwing.com/problem/content/274/

算法

• 状态表示：\(f[i, j]\)
  • 集合：所有由第一个序列的前 \(i\) 个字母和第二个序列前 \(j\) 个字母构成的，且以 \(b[j]\) 结尾的公共上升子序列
  • 属性：Max
• 状态计算
  • 所有包含 \(a[i]\) 的公共子序列: \(f[i, k] + 1\)
  • 所有不包含 \(a[i]\) 的公共子序列: \(f[i - 1, j]\)

暴力代码

#include <iostream>
#include <algorithm>
using namespace std;

const int N = 3010;
int n;
int a[N], b[N], f[N][N];

int main() {
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 1; i <= n; ++i) cin >> b[i];
    
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            f[i][j] = f[i - 1][j];
            if (a[i] == b[j]) {
                f[i][j] = max(f[i][j], 1);
                for (int k = 1; k < j; ++k) {
                    if (b[k] < b[j]) f[i][j] = max(f[i][j], f[i][k] + 1);
                }
            }
        }
    }
    
    int ans = 0;
    for (int i = 1; i <= n; ++i) ans = max(ans, f[n][i]);
    cout << ans << endl;
    
    return 0;
}

优化代码

#include <iostream>
#include <algorithm>
using namespace std;

const int N = 3010;
int n;
int a[N], b[N], f[N][N];

int main() {
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 1; i <= n; ++i) cin >> b[i];
    
    for (int i = 1; i <= n; ++i) {
        int maxv = 1;
        for (int j = 1; j <= n; ++j) {
            f[i][j] = f[i - 1][j];
            if (a[i] == b[j]) f[i][j] = max(f[i][j], maxv);
            if (a[i] > b[j]) maxv = max(f[i][j] + 1, maxv);
        }
    }
    
    int ans = 0;
    for (int i = 1; i <= n; ++i) ans = max(ans, f[n][i]);
    cout << ans << endl;
    
    return 0;
}