Common Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28988 Accepted Submission(s): 12983
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
Recommend
一道dp 的题，求两个字符串的最大公共子序列，第一次写，没有思路，看了别人的代码，感觉还是比较容易理解的。
题意：求两个字符串的最大公共子序列长度，可以不连续。
附上代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #define N 1001 //N不要取太多，超了一次内存 5 using namespace std; 6 char a[N],b[N]; 7 int dp[N][N]; 8 int max(int a,int b) //取最大值 9 { 10 return a>b?a:b; 11 } 12 int main() 13 { 14 int n,m,i,j; 15 while(~scanf("%s%s",a,b)) 16 { 17 n=strlen(a); //取两个序列的长度 18 m=strlen(b); 19 memset(dp,0,sizeof(dp)); 20 for(i=1; i<=n; i++) 21 for(j=1; j<=m; j++) 22 { 23 if(a[i-1]==b[j-1]) 24 dp[i][j]=dp[i-1][j-1]+1; //若相等，则在dp[i-1][j-1]基础上加1 25 else 26 dp[i][j]=max(dp[i-1][j],dp[i][j-1]); //若不相等，则选择两个字符串中有相同公共子序列多个记录下来 27 } 28 printf("%d ",dp[n][m]); 29 } 30 return 0; 31 }