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; }