杭电1159——Common Subsequence

Problem Description

A subsequence of a given sequence is the given sequence withsome elements (possible none) left out. Given a sequence X =<x1, x2, ..., xm> another sequence Z= <z1, z2, ..., zk> is a subsequenceof X if there exists a strictly increasing sequence<i1, i2, ..., ik> of indices of Xsuch 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 indexsequence <1, 2, 4, 6>. Given twosequences X and Y the problem is to find the length of themaximum-length common subsequence of X and Y.
The program input is from a text file. Each data set in the filecontains two strings representing the given sequences. Thesequences are separated by any number of white spaces. The inputdata are correct. For each set of data the program prints on thestandard output the length of the maximum-length common subsequencefrom the beginning of a separate line.

 

Sample Input

abcfbc abfcab
programming contest
abcd mnp

Sample Output

4 2 0

# include<stdio.h>

# include<string.h>

int num[1001][1001];

int main()

{

    int i,j,len1,len2,a,b;

    char s1[1000],s2[1000];

    while(scanf("%s%s",s1+1,s2+1)!=EOF)

    {

        len1=strlen(s1+1);

        len2=strlen(s2+1);

        for(i=0;i<len1+1;i++)

            num[i][0]=0;

        for(i=0;i<len2+1;i++)

            num[0][i]=0;

        for(i=1;i<=len1;i++)

            for(j=1;j<=len2;j++)

            {

                if(s1[i]==s2[j])

                  num[i][j]=num[i-1][j-1]+1;

                 else

                 {

                     a=num[i][j-1];

                     b=num[i-1][j];

                     if(a>b)

                         num[i][j]=a;

                     else

                         num[i][j]=b;

                 }

            }

        printf("%d\n",num[len1][len2]);

    }

    return 0;

}