1、最大公共子串
dp[i][j] 表示 s1 以 i 结尾和 s2 以 j 结尾的最大公共子串
#include <iostream> #include<vector> #include<algorithm> using namespace std; int main() { string s1, s2; cin >> s1 >> s2; int m = s1.size(), n = s2.size(); int ans = 0; vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0)); for(int i = 1; i <= m; i ++) for (int j = 1; j <= n; j++) { if (s1[i - 1] == s2[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1; else dp[i][j] = 0; ans = max(ans, dp[i][j]); } cout << ans << endl; return 0; }
最长公共子序列
int longestCommonSubsequence(string text1, string text2) { int m = text1.size(), n = text2.size(); vector<vector<int>>dp(m + 1, vector<int>(n + 1, 0)); for(int i = 1; i <= m; i ++){ for(int j = 1; j <= n; j ++){ if(text1[i - 1] == text2[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1; else dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]); } } return dp[m][n]; }
2、后缀求值
#include <iostream> #include <stack> using namespace std; int main(){ int temp1, temp2; char cur; string str; getline(cin, str); int i = 0; stack<int> st; while (i < str.size()){ cur = str[i]; if (cur >= '0' && cur <= '9') st.push(cur - '0'); else { temp2 = st.top(); st.pop(); temp1 = st.top(); st.pop(); if (cur == '+') st.push(temp1 + temp2); else if (cur == '-') st.push(temp1 - temp2); } i++; } cout << st.top(); return 0; }
3、哈夫曼编码
#include <iostream> #include<queue> #include<algorithm> using namespace std; int main() { int n; cin >> n; priority_queue<int, vector<int>, greater<int>> q; int sum = 0, num; for (int i = 0; i < n; i++) { cin >> num; q.push(num); } int num1, num2; while (q.size() > 1) { num1 = q.top(); q.pop(); num2 = q.top(); q.pop(); sum += num1 + num2; q.push(num1 + num2); } cout << sum << endl; return 0; }