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RAID磁盘阵列

2024-07-06 19:35:00   231人阅读

 　　　　　　　　　　　　　　　　　　　　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 }

 

 

 


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