[POJ] 1458 -> Common Subsequence

Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 41382		Accepted: 16722

DescriptionA 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 strictlyincreasing 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 Xand Y the problem is to find the length of the maximum-length common subsequence of X and Y.InputThe 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.OutputFor 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

解题思路：

Code：

#include <iostream>using namespace std;#define MAX_SIZE 256int C[MAX_SIZE][MAX_SIZE];inline int LCS(const char* str1, const char* str2){int len1 = strlen(str1);int len2 = strlen(str2);for (int i = 0; i < len1; ++i){for (int j = 0; j < len2; ++j){if (i == 0 || j == 0) C[i][j] = 0;else if (str1[i] == str2[j]) C[i][j] = C[i - 1][j - 1] + 1;else C[i][j] = (C[i - 1][j] > C[i][j - 1]) ? C[i - 1][j] : C[i][j - 1];}}return C[len1 - 1][len2 - 1];}int main(){char a[256], b[256];while (cin >> a+1 >> b+1)cout << LCS(a, b) << endl;return 0;}