poj 1458 Common Subsequence

Common Subsequence

Time Limit: 1000MSMemory Limit:10000K

Total Submissions: 26378Accepted:10221

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, 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 abfcab programming contest abcd mnp

Sample Output

4 2 0   #include <iostream> #include <string> using namespace std; string str1; string str2; int num[256][256]; int lcs ( string s1 , string s2 , int m , int n ) { int i, j; for ( i=1 ; i <= m ; i++ ) { for ( j = 1 ; j <= n ; j++ ) { if ( s1[i-1] == s2[j-1] ) { num[i][j] = num[i-1][j-1] + 1; } else { if ( num[i][j-1] < num[i-1][j] ) { num[i][j] = num[i-1][j]; } else { num[i][j] = num[i][j-1]; } } } } return num[m][n]; } int main() { int len1, len2; while ( cin>>str1>>str2 ) { memset ( num , 0 , sizeof(num) ); len1 = str1.length(); len2 = str2.length(); int out = lcs(str1,str2,len1,len2); cout<<out<<endl; } return 0; }

转载于:https://www.cnblogs.com/w0w0/archive/2011/11/21/2257538.html

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