POJ1458 Common Subsequence

URL : http://poj.org/problem?id=1458

      LCS 的水题，做另外一个动态规划题没思路，就找到了它放松下脑子，代码：

#include <cstdio>#include <iostream>#include <string>using namespace std;int main(){    string s1, s2;    while(cin >> s1 >>s2){         int m = s1.size(), n = s2.size();         int LCS[m+1][n+1];         for(int i = 0; i <= n; i++)            LCS[0][i] = 0;         for(int j = 0; j <= m; j++)            LCS[j][0] = 0;         for(int i = 1; i <= m; i++)             for(int j = 1; j <= n; j++)                 if(s1[i-1] == s2[j-1])                     LCS[i][j] = LCS[i-1][j-1] + 1;                 else                     LCS[i][j] = LCS[i][j-1] > LCS[i-1][j] ? LCS[i][j-1] : LCS[i-1][j];         printf("%d\n", LCS[m][n]);    }    return 0;}