Problem Description
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, xij = 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.
The program input is from a text file. Each data set in the file contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct. 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 abfcab
programming
contest abcd mnp

Sample Output
4 2 0
先找第一串第一個字符與第二串的長度， 再找第一串前兩個字符與第二串的最長公共子序列長度， 以此類推。 以abcfb 與 abfcab為例，附空間輔助變化示意圖，求這兩串的最長公共子序列 圖中設置的是二維數組，格內的數為行列號，比如，第一個00表示第0行第0列， 由圖可以很清晰的看出表達式：F[i][j]=F[i-1][j-1]+1；（a[i]==b[j]）F[i][j]=max(F[i-1][j],F[i][j-1])(a[i]!=b[j]); 當a[i]==b[j]時，長度為 二維數組[i-1][j-1]+1， 不相等時為上格子與前一格子最大值。

代碼如下

#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int maxn=1000; int dp[maxn][maxn]; int main() {string x,z;int max1,lx,lz;while(cin>>x>>z){memset(dp,0,sizeof(dp));max1=0;lx=x.size(),lz=z.size();for(int i=1;i<=lx;i++){for(int t=1;t<=lz;t++){dp[i][t]=max(dp[i][t-1],dp[i-1][t]);if(x[i-1]==z[t-1]) dp[i][t]=dp[i-1][t-1]+1;if(dp[i][t]>max1) max1=dp[i][t];}}cout<<max1<<endl;}return 0; }

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