AtCoder · dp_f

F - LCS

This C++ solution uses dynamic programming for AtCoder dp_f F - LCS. Read the reasoning, inspect the code, or try your own test case below.

dp_fDynamic programmingC++35 lines
Solution126of 248
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Approach

Dynamic programming

Educational DP Contest LCS task requires reconstructing one longest common subsequence; the DP and backtracking match.

Dynamic programming

Problem and code

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Written by benbenyaojifen. Try the problem first, then compare your approach with the code.

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Implementation

LCS.cpp

C++

    #include <bits/stdc++.h>
    using namespace std;
    using ll = long long;
    using i128 = __int128;
    const int inf = 1e9;
    const ll INF = 1e18; //❄️
    int main() {
        ios::sync_with_stdio(0); cin.tie(0); 
        string a, b; cin >> a >> b;
        vector<vector<int>> dp(a.size() + 1, vector<int>(b.size() + 1));
        for (int i = 1; i <= a.size(); i++) {
            for (int j = 1; j <= b.size(); j++) {
                if (a[i - 1] == b[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1;
                else dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        auto out = [&](const vector<vector<int>> &dp) {
            string ans = "";
            int i = a.size(), j = b.size();
            while (i != 0 && j != 0) {
                if (dp[i][j] == dp[i - 1][j - 1] + 1 && a[i - 1] == b[j - 1]) {
                    i--; j--; ans += a[i];
                } else if (dp[i][j] == dp[i - 1][j]) {
                    i--;
                } else {
                    j--;
                }
            }
            return ans;
        };
        string s = out(dp);
        reverse(s.begin(), s.end());
        cout << s << '\n';
        return 0;
    }
        

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