Common Subsequence dp

A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence Z = < z1, z2, ..., zk > is a subsequence of X if there exists a strictly increasing sequence < i1, i2, ..., ik > of indices of X such that for all j = 1,2,...,k, x _ij = zj. For example, Z = < a, b, f, c > is a subsequence of X = < a, b, c, f, b, c > with index sequence < 1, 2, 4, 6 >. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.

Input

The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.

Output

For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.

Sample Input

abcfbc         abfcabprogramming    contest abcd           mnp

Sample Output

420

题意：给定两个字符串，求其最长公共子串。其中子串不必是连续的，即ac是abc的子串。

用dp[i][j]来表示a串前i个字符与b串前j个字符的公共子串最长长度。

状态转移方程：

如果a[i]==b[j]  dp[i][j]=dp[i-1][j-1]+1; 因为a[i]和b[j]是相等的，所以长度就是dp[i-1][j-1]加一就行

如果a[i]!=b[j]  dp[i][j]=max(dp[i-1][j],dp[i][j-1]); 因为a[i]和b[j]是不相等的，所以现在的最长就是a的前i-1在b的前j的最长和a的前i在b的前j-1的最长中的较大值。

#include<iostream>#include<cstring>#include<cmath>#include<algorithm>using namespace std;int dp[1000][1000];int main() {string a,b;while(cin>>a>>b) {int La,Lb;La=a.size() ;Lb=b.size() ;int Max=0;memset(dp,0,sizeof(dp));for(int i=0; i<a.size(); i++) {for(int j=0; j<b.size(); j++) {if(a[i]==b[j]) {if(i>0&&j>0)dp[i][j]=dp[i-1][j-1]+1;elsedp[i][j]=1;}if(a[i]!=b[j]) {if(i>0&&j>0)dp[i][j]=max(dp[i-1][j],dp[i][j-1]);else if(i>0)dp[i][j]=dp[i-1][j];else if(j>0)dp[i][j]=dp[i][j-1];}Max=max(Max,dp[i][j]);}}cout<<Max<<endl;}}