Common Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30912 Accepted Submission(s): 13987
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
Recommend
Ignatius
L[i+1][j+1]=max(L[i][j+1],L[i+1][j]); L[i][j]表示字符串A的1~i+1子串与字符串B的1~j+1子串的公共子串的长度。
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 using namespace std;
5
6 const int maxn=1050;
7 int L[maxn][maxn];
8 char s1[maxn];
9 char s2[maxn];
10
11 int main()
12 {
13 int len1,len2;
14 while(scanf("%s%s",s1,s2)==2)
15 {
16 len1=strlen(s1);
17 len2=strlen(s2);
18 for(int i=0;i<len1;i++)
19 for(int j=0;j<len2;j++)
20 L[i][j]=0;
21 for(int i=0;i<len1;i++)
22 {
23 for(int j=0;j<len2;j++)
24 if(s1[i]==s2[j]) L[i+1][j+1]=L[i][j]+1;
25 else L[i+1][j+1]=max(L[i][j+1],L[i+1][j]);
26 }
27 printf("%d\n",L[len1][len2]);
28 }
29 }
HDU 1159 Common Subsequence
原文：http://www.cnblogs.com/cumulonimbus/p/5185085.html
