ZOJ 1733(Common Subsequence)

//经典dp
//最长公共子序列问题

#include<iostream>#include<algorithm>#include<string>using namespace std;//dp[i][j]表示的是 s1[0],s1[1],s1[2]....s1[i-1](共i个)和s2[0],s2[1],s2[2]....s2[j-1]（共j个）的最长公共子序列的长度int dp[1001][1001];  //数组记得开大点int main(void){    string s1, s2;    int len1, len2;    while (cin >> s1 >> s2)    {        len1 = s1.size(), len2 = s2.size();        //以下两段代码也可改成 memset(dp,0,sizeof(dp));        for (int i = 0; i <= len1; i++) dp[i][0] = 0;        for (int i = 0; i <= len2; i++) dp[0][i] = 0;        for (int i = 1; i <= len1; i++)        {            for (int j = 1; j <= len2; j++)            {                if (s1[i - 1] == s2[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1;                else dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);            }        }        cout << dp[len1][len2] << endl;    }    return 0;}