POJ 1458 （简单LCS）

╰+哭是因爲堅強的太久メ 2024-02-17 19:26 38阅读 0赞

Common Subsequence

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;50110</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;20690</td> 
  </tr> 
 </tbody> 
</table>

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

AC代码：

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int maxn=1000;
    int dp[maxn][maxn];
    char s1[maxn],s2[maxn];
    int main(){
    	while(scanf("%s%s",s1,s2)==2){
    		int len1=strlen(s1);
    		int len2=strlen(s2);
    		memset(dp,0,sizeof(dp));
    		
    		for(int i=1;i<=len1;i++)
    		 for(int 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;
    }