Common Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38176 Accepted Submission(s): 17504

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.

Sample Input

abcfbc abfcab

programming contest

abcd mnp

Sample Output

Source

Southeastern Europe 2003

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动态规划中的最长公共子序列。用dp[i][j]保存第一个字符串前i个字符和第二个字符串前j个字符的公共子序列长度。

如果str1[i]==str2[j]，dp[i][j]=dp[i-1][j-1]。

如果str1[i]!=str2[j]，dp[i][j]=max(dp[i-1][j],dp[i][j-1])

上述即是动态规划的状态转移公式。

 1 #include <stdio.h>
 2 #include <string.h>
 3 
 4 using namespace std;
 5 
 6 int dp[1001][1001];
 7 
 8 char str1[1001], str2[1001];
 9 
10 int max(int a, int b)
11 {
12     if(a>b)
13         return a;
14     else
15         return b;
16 }
17 
18 int main()
19 {
20     while(scanf("%s %s", str1, str2)!=EOF)
21     {
22 
23         int len1=strlen(str1);
24         int len2=strlen(str2);
25 
26 
27         for(int i=0; i<=len1; i++) dp[i][0]=0;
28         for(int j=0; j<=len2; j++) dp[0][j]=0;
29 
30         for(int i=1; i<=len1; i++)
31             for(int j=1; j<=len2; j++)
32         {
33             if(str1[i-1]!=str2[j-1])
34             {
35                 dp[i][j]=max(dp[i-1][j], dp[i][j-1]);
36 
37             }
38 
39             else
40             {
41                 dp[i][j]=dp[i-1][j-1]+1;
42             }
43         }
44 
45     printf("%d
",dp[len1][len2]);
46     }
47 
48     return 0;
49 }