Common Subsequence
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.
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 abfcabprogramming contest abcd mnp
Sample Output
420
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
char s1[1005],s2[1005];
int dp[1005][1005];
int i,j,len1,len2;
while(scanf("%s%s",s1,s2)!=EOF)
{
len1=strlen(s1);
len2=strlen(s2);
memset(dp,0,sizeof(dp));
for(i=1;i<=len1;i++)
{
for(j=1;j<=len2;j++)
{
if(s1[i-1]==s2[j-1])
dp[i][j]=dp[i-1][j-1]+1;
else
dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
}
}
printf("%d\n",dp[len1][len2]);
}
return 0;
}
本文介绍了一种解决最长公共子序列问题的高效算法。通过动态规划方法,该程序能够找出两个字符串之间的最大长度公共子序列。输入为两组字符串,输出则是它们共同子序列的最大长度。
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