Common Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 39087 Accepted Submission(s): 17978
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 abfcab
programming contest
abcd mnp
Sample Output
4
2
0
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[1005][1005];
int main(){
char c1[1005],c2[1005];
while(~scanf("%s %s",c1+1,c2+1)){
getchar();
int len1 = strlen(c1+1);
int len2 = strlen(c2+1);
for(int i=1;i<=len1;i++){
for(int j=1;j<=len2;j++){
if(c1[i]==c2[j])dp[i][j] = dp[i-1][j-1]+1;
else dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
}
}
cout<<dp[len1][len2]<<endl;
}
return 0;
}
本文介绍了一个经典的计算机科学问题——寻找两个序列之间的最长公共子序列,并提供了一段使用动态规划解决该问题的C++代码实现。通过输入两个字符串,程序能够计算并输出最长公共子序列的长度。
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