Common Subsequence

abcfbc abfcab
programming contest 
abcd mnp

4
2
0

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<map>

#define N 1010
#define Mod 10000007
#define lson l,mid,idx<<1
#define rson mid+1,r,idx<<1|1
#define lc idx<<1
#define rc idx<<1|1
const double EPS = 1e-11;
const double PI = acos ( -1.0 );
const double E = 2.718281828;
typedef long long ll;

const int INF = 1000010;

using namespace std;

char a[N],b[N];
int dp[N][N];

int main() {
    while(~scanf("%s%s",a,b)) {
        int l1=strlen(a);
        int l2=strlen(b);
        memset(dp,0,sizeof dp);
        //dp[0][0]=1;
        for(int i=0; i<l1; i++) {
            for(int j=0; j<l2; j++) {
                if(a[i]==b[j])dp[1+i][1+j]=dp[i][j]+1;
                else
                    dp[i+1][1+j]=max(dp[i+1][j],dp[i][j+1]);
            }
        }
        printf("%d\n",dp[l1][l2]);
    }
    return 0;
}

hdu 1159 Common Subsequence(dp)

原文：http://blog.csdn.net/acm_baihuzi/article/details/44502819