Common Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32693 Accepted Submission(s): 14786
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.
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
4
2
0
Source
题意：求两个字符串的最长公共子序列的长度
题解： dp[i][j] 表示a串前i个字符与b串前j个字符 最长公共子序列的长度 N(n*m)
当a[i]==a[j]时 dp[i][j]=dp[i-1][j-1]+1;
否则 dp[i][j]=max（dp[i-1][j],dp[i][j-1]）
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 using namespace std; 5 char a[1005],b[1005]; 6 int dp[1005][1005]; 7 int main() 8 { 9 while(cin>>a>>b) 10 { 11 int len1=strlen(a); 12 int len2=strlen(b); 13 memset(dp,0,sizeof(dp)); 14 for(int i=1;i<=len1;i++) 15 for(int j=1;j<=len2;j++) 16 { 17 if(a[i-1]==b[j-1]) 18 dp[i][j]=dp[i-1][j-1]+1; 19 else 20 dp[i][j]=max(dp[i-1][j],dp[i][j-1]); 21 } 22 cout<<dp[len1][len2]<<endl; 23 } 24 return 0; 25 }