Common Subsequence
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 36542 |
|
Accepted: 14596 |
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.
Input
The program input is from the std input. Each data
set in the input contains two strings representing the given sequences. The
sequences are separated by any number of white spaces. The input data are
correct.
Output
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
入门dp，状态转移方程： if(a[j] == b[i]) dp[i][j] = dp[i-1][j-1] + 1；
else dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
就是说如果a[j]=b[i]，那么lcs就在前面状态的基础上加一，如果不成立，可以从两个方向考虑lcs，不解释~~~
AC代码如下：
1 #include <cstdio>
2 #include <string.h>
3 #include <iostream>
4
5 using namespace std;
6
7 char a[1000];
8 char b[1000];
9 int dp[1000][1000];
10
11 int main()
12 {
13 int m, n;
14 while(scanf("%s%s", a, b) != EOF) {
15 m = strlen(a);
16 n = strlen(b);
17 for(int i = 1; i <= m; i++) {
18 for(int j = 1; j <=n+1; j++) {
19 if(b[j-1] == a[i-1])
20 dp[i][j] = dp[i-1][j-1] + 1;
21 else
22 dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
23 }
24 }
25 printf("%d\n", dp[m][n]);
26 }
27 return 0;
28 }
RAID磁盘阵列,布布扣,bubuko.com
RAID磁盘阵列
原文：http://www.cnblogs.com/oracle-dba/p/3763321.html
