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POJ 1458 Common Subsequence 【最长公共子序列】



Common Subsequence


Time Limit: 1000MS

 

Memory Limit: 10000K

Total Submissions: 39741

 

Accepted: 15979


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, x ij = 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.


Input


The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.


Output


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



dp方程:
    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])

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
int dp[1010][1010];
int main()
{
int i,j;
string s1,s2;
while(cin>>s1>>s2)
{
memset(dp,0,sizeof(dp));
for(i=1; i<=s1.length(); i++)
{
for(j=1; j<=s2.length(); 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[s1.length()][s2.length()]);
}
}






























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