HDOJ 题目1159 Common Subsequence(LCS)

2015-07-20 17:06:53 ? 作者: ? 浏览: 3

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25791 Accepted Submission(s): 11432



Problem Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = another sequence Z = is a subsequence of X if there exists a strictly increasing sequence of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = is a subsequence of X = 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.

Sample Input
abcfbc abfcab
programming contest 
abcd mnp

Sample Output
4
2
0

Source Southeastern Europe 2003
Recommend Ignatius | We have carefully selected several similar problems for you: 2577 1074 1158 1080 1078 ac代码
#include
       
        
#include
        
          #define max(a,b) (a>b?a:b) int dp[1010][1010],len1,len2; char s1[1010],s2[1010]; void LCS() { int i,j; 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]); } } } int main() { while(scanf("%s%s",&s1,s2)!=EOF) { len1=strlen(s1); len2=strlen(s2); LCS(); printf("%d\n",dp[len1][len2]); } }
        
       


-->

评论

帐  号: 密码: (新用户注册)
验 证 码:
表  情:
内  容: