题解 | #最长公共子序列(一)#

动态规划板子
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

int dp[1002][1002];

int main() {
    int n, m;
    char s1[1001];
    char s2[1001];
    scanf("%d%d", &n, &m);
    scanf("%s%s", s1 + 1, s2 + 1);

    // dp[i][j] s1的前i个元素 s2的前j个元素 的最大公共长度
    for (int j = 0; j <= m; ++j) {
        dp[0][j] = 0;
    }
    for (int i = 0; i <= n; i++) {
        dp[i][0] = 0;
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            // s1[1]- s1[i] 和 s2[1]-s2[j]
            if (s1[i] == s2[j]) {
                dp[i][j] = dp[i - 1][j - 1] + 1;
            } else {
                dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]);
            }

        }
    }
    cout << dp[n][m] << endl;
    return 0;
}