poj-1485 Common Subsequence

基础dp的题目；

最长公共子序列，，，LCS(Longest Common Subsequence)

dp[i][j]表示分别以i和j结尾的的最长公共子序列的长度；

如果s1[i]==s2[j] dp[i][j]就等于dp[i-1][j-1]的值加上一；

如果是s1[i]!=s2[j] dp[i][j]应该等于dp[i-1][j]和dp[i][j-1]的最大值

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

const int maxn=1e3+5;
char a[maxn],b[maxn];
int dp[maxn][maxn];

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