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;
} 
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