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最新推荐文章于 2018-11-18 14:42:38 发布

sonfree

最新推荐文章于 2018-11-18 14:42:38 发布

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分类专栏： ACM学习笔记文章标签： output input each zk file c

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本文探讨了如何通过输入两段字符串，计算它们的最大公共子序列长度，详细介绍了算法实现和示例输入输出。

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Common Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10295 Accepted Submission(s): 4237

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

#include<stdio.h>
#include<string.h>
main()
{
    int a[501][501]={0},i,j,m,n;
    char ch1[501],ch2[501];
    while(scanf("%s%s",ch1,ch2)!=EOF)
    {
        m=strlen(ch1);
        n=strlen(ch2);
        for(i=0;i<m;i++)
        {
            for(j=0;j<n;j++)
            {
                if(ch1[i]==ch2[j]) a[i+1][j+1]=a[i][j]+1;
                else if(a[i][j+1]>=a[i+1][j]) a[i+1][j+1]=a[i][j+1];
                else if(a[i][j+1]<a[i+1][j]) a[i+1][j+1]=a[i+1][j];
            }
        }
        printf("%d\n",a[i][j]);
    }

}