Common Subsequence
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 34477 | Accepted: 13631 |
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
Source
Southeastern Europe 2003 题意： 求两个字符串的最长公共子序列(lcs)的长度。 思路： 金典问题，动态转移方程如下：
1 if（i==0||j==0）
2 dp[i][j]=0;
3 else if(x[i]==y[i])
4 dp[i][j]=dp[i-1][j-1]+1;
5 else
6 dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
代码如下：
1 #include<iostream>
2 #include<string>
3 #include<cmath>
4 using namespace std;
5 const int MAX=500;
6 int dp[MAX][MAX]={0};
7 int max(int a,int b)
8 {
9 return a>b?a:b;
10 }
11 int main(int *argv,int *argc[])
12 {
13 int len1,len2,i,j;
14 string str1,str2; //有时在时间允许范围内，用cin读字符串比较方便
15 while(cin>>str1>>str2)
16 {
17 len1=str1.length();
18 len2=str2.length();
19 for(i=1;i<=len1;i++)
20 {
21 for(j=1;j<=len2;j++)
22 //转移方程
23 {
24 if(str1[i-1]==str2[j-1])
25 dp[i][j]=dp[i-1][j-1]+1;
26 else
27 dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
28 }
29 }
30 cout<<dp[len1][len2]<<endl;
31 }
32 return 0;
33 }
View Code
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