zoj 1733 Common Subsequence

/*动态规划题也许是刚刚接触动态规划的缘故发现他真的是奇妙无穷许多看似无法可解的问题用它就可以很轻易地解决此题是求最长公共子串的问题状态转移方程dp[i][j]=dp[i-1][j-1]+1; (str1[i]==str2[j])dp[i][j]=MAX{dp[i-1][j],dp[i][j-1]}; (str1[i]!=str2[j])最后输出dp[i][j]即可对于字符串str1="X1X2...Xi...Xm";str2="Y1Y2...Yj...Yn";若Xi==Yj则str1，str2的最长公共子序列长一定等于"X1X2...Xi-1" ,"Y1Y2...Yj-1"的最长公共子序列长度+1同理若Xi！=Xj，则str1，str2的最长公共子序列长一定等于"X1X2...Xi-1"和"Y1Y2...Yj" ,"Y1Y2...Yj-1"和"X1X2...Xi"中较长的公共子序列长度*/#define LOCAL#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<cstdlib>#include<iomanip>#include<string>#include<algorithm>#include<ctime>#include<stack>#include<queue>#include<vector>#define N 1005using namespace std;int dp[N][N];int main(){#ifdef LOCAL       freopen("input.txt","r",stdin);       freopen("output.txt","w",stdout);#endifstring str1,str2;int strlen1,strlen2,i,j;while(cin>>str1>>str2){memset(dp,0,sizeof(dp));strlen1=str1.size();strlen2=str2.size();for(i=1;i<=strlen1;i++){for(j=1;j<=strlen2;j++){if(str1[i-1]==str2[j-1]) dp[i][j]=dp[i-1][j-1]+1;else dp[i][j]=(dp[i-1][j]>dp[i][j-1]?dp[i-1][j]:dp[i][j-1]);}    }cout<<dp[strlen1][strlen2]<<endl;}return 0;}