Common Subsequence （最长公共子序列）

HDU - 1159

                         Common Subsequence      HDU - 1159 

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. 

Input

abcfbc abfcab programming contest abcd mnp

Output

4 2 0

Sample Input

abcfbc abfcab programming contest abcd mnp

Sample Output

4 2 0

题目大意：给出两个字符串，求两个字符串的最长公共字串。

思路：最长的序列一定是由前边的最长子序列的基础上确定的，因此最长子序列 dp[i][j] 可以由前边的的推导得来dp[i-1][j-1]+1;

#include <string> #include <iostream> #include <cmath> #include <algorithm> using namespace std; const int maxn=5005; int dp[maxn][maxn]; int main() { string c1; string c2; while(cin>>c1>>c2) { int len1=c1.size(); int len2=c2.size(); // cout <<len1<<" "<<len2<<endl; for(int i=1; i<=len1; i++) { for(int j=1; j<=len2; j++) { if(c1[i-1]==c2[j-1]) dp[i][j]=dp[i-1][j-1]+1; else dp[i][j]=max(dp[i-1][j],dp[i][j-1]); // cout <<dp[i][j]<<" "; } // cout <<endl; } cout <<dp[len1][len2]<<endl; } return 0; }