hdu 1159 经典dp最长公共子序列

dp[i][j]=max{max[i-1][j],max[i][j-1],max[i-1][j-1]+(a[i] == b[j])}
//dp[i][j]定以为，a串的前i个字符和b串的前b个字符的最大字串和，为选a串的第i个字符而不选b串的第j个字符，不选a串的第i个字符而选b串的第j个字符，既选a串的第i个字符又选b串的第j个字符三者中较大的

#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
using namespace std;

int main(void){
    string a,b;
    while(cin >> a) {
        cin >> b;
        int n = a.size(), m = b.size();
        int dp[n + 9][m + 9];
        memset(dp,0,sizeof(dp));
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                dp[i][j] = max(max(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1] + (a[i-1] == b[j-1]));
            }
        }
        cout << dp
[m] << endl;
    }
    return 0;
}