标签：out arch represent ace logs from ret min lcs问题
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
Source
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：
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）
标签：out arch represent ace logs from ret min lcs问题
原文地址：http://www.cnblogs.com/zmin/p/6858816.html
