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.
Input
abcfbc abfcab
programming contest
abcd mnp
Output
4
2
0
Sample Input
abcfbc abfcab
programming contest
abcd mnp
Sample Output
4
2
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.
Input
abcfbc abfcab
programming contest
abcd mnp
Output
4
2
0
Sample Input
abcfbc abfcab
programming contest
abcd mnp
Sample Output
4
2
0
最大公共子序列。dp[i][j]表示前i个字符与前j个字符相同的个数
AC代码:
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<string>
#include<queue>
#include<set>
#include<vector>
#include<map>
#include<stack>
#include<cstdlib>
using namespace std;
typedef long long ll;
char a[1005],b[1005];
int dp[1005][1055];
int main()
{
int ans;
while(scanf("%s%s",a,b)!=EOF){
memset(dp,0,sizeof(dp));
int la=strlen(a);
int lb=strlen(b);
// printf("%d %d\n",la,lb);
for(int i=0;i<la;i++)
dp[i][0]=0;
for(int i=0;i<lb;i++)
dp[0][i]=0;
for(int i=1;i<=la;i++)
for(int j=1;j<=lb;j++){
if(a[i-1]==b[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 ",ans);
}
printf("%d\n",dp[la][lb]);
}
return 0;
}
最长公共子序列求解
本文介绍了一种解决最长公共子序列问题的有效算法,并通过动态规划实现。输入为两个字符串,输出则是这两个字符串间的最长公共子序列的长度。
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