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 abfcabprogramming contest abcd mnp
Sample Output
420
#include#include #include #include #include using namespace std;#define met(a,b) (memset(a,b,sizeof(a)))#define N 1100#define INF 0xffffffchar s1[N], s2[N];int dp[N][N];int main(){ while(scanf("%s%s", s1, s2)!=EOF) { int i, j, len1=strlen(s1), len2=strlen(s2); met(dp, 0); ///要考虑下下标越界的问题 for(i=1; i<=len1; i++) for(j=1; j<=len2; j++) { if(s1[i-1]==s2[j-1]) dp[i][j] = dp[i-1][j-1] + 1; else dp[i][j] = max(dp[i-1][j], dp[i][j-1]); } printf("%d\n", dp[len1][len2]); } return 0;}