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hdu-1159 Common Subsequence （dp中的lcs问题）

2024-09-22 11:45:28   207人阅读

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38425    Accepted Submission(s): 17634


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
4 2 0
 

 

Source
Southeastern Europe 2003
 
 
本题就是经典的lcs问题哈，还是老套路，把一个问题分解为若干小问题。
得到如下递推式：
 
技术分享
下面附代码：
 1 #include<cstdio>
 2 #include<cstring>
 3 #include<cmath>
 4 #include<algorithm>
 5 using namespace std;
 6 const int Max = 1111;
 7 char st1[Max],st2[Max];
 8 int dp[Max][Max];
 9 int main()
10 {
11     while(~scanf("%s %s",st1,st2))
12     {
13         memset(dp,0,sizeof(dp));
14         int m=strlen(st1);
15         int n=strlen(st2);
16         for(int i=0;i<m;i++)
17         for(int k=0;k<n;k++)
18         {
19             if(st1[i]==st2[k])
20             dp[i+1][k+1]=dp[i][k]+1;
21             else
22             dp[i+1][k+1]=max(dp[i+1][k],dp[i][k+1]);
23         }
24         printf("%d\n",dp[m][n]);
25     }
26     return 0;
27 }

 

hdu-1159 Common Subsequence （dp中的lcs问题）


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