05_最长公共子序列问题(LCS)

问题来源：刘汝佳《算法竞赛入门经典--训练指南》 P60 问题7：

问题描述：给两个子序列A和B，求长度最大的公共子序列。比如1,5,2,6,8,和2,3,5,6,9,8,4的最长公共子序列为5,6,8另一个解是2,6,8)。

分析：设dp[i][j]为A1,A2,...,Ai和B1,B2,...,Bn的LCS长度，则状态转移方程为：

1 if(A[i]==B[i])　　 2 　　d[i][j] = d[i-1][j-1]+1; 3 else

4 　　d[i][j] = Max{d[i-1][j],d[i][j-1]};

时间复杂度为O(n*m);其中空间可用滚动数组优化。

例题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1159

例题：hdu 1159

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25784    Accepted Submission(s): 11428

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

 

题意：给两个字符串，求两字符串的LCS

代码：

 1 #include "stdio.h"

 2 #include "string.h"

 3 #define N 1005

 4 

 5 int dp[N];

 6 char s1[N],s2[N];

 7 

 8 int inline Max(int a,int b) { return a>b?a:b; }

 9 

10 int main()

11 {

12     int i,j;

13     int pre,next;

14     int len1,len2;

15     while(scanf("%s %s",s1,s2)!=EOF)

16     {

17         len1 = strlen(s1);

18         len2 = strlen(s2);

19         memset(dp,0,sizeof(dp));

20         for(i=0; i<len1; i++)

21         {

22             next = 0;

23             for(j=0; j<len2; j++)

24             {

25                 pre = dp[j];  //pre和next将下一次要用到的dp[i-1][j-1]先存起来，实现空间压缩

26                 if(s1[i]==s2[j])

27                     dp[j] = next+1;

28                 else

29                     dp[j] = Max(dp[j-1],dp[j]);

30                 next = pre;

31             }

32         }

33         printf("%d\n",dp[len2-1]);

34     }

35     return 0;

36 }