Common Subsequence
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
题目类型:线性DP
算法分析:求最大公共子串的长度,f[i][j]定义为有前i和j个字符的字符串val_a和val_b的最长公共子串的长度。转移方程为:如果字符串val_a[i] == val_b[j],则有f[i][j] = f[i-1][j-1] + 1;反之则f[i][j] = max (f[i-1][j], f[i][j-1]); 1 <= i, j
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#include <iostream> #include <fstream> #include <algorithm> #include <iomanip> #include <cstring> #include <cstdio> #include <cmath> #include <map> #include <string> #include <vector> #include <stack> #include <queue> #include <set> #include <list> #include <ctime> using namespace std; const int maxn = 266; const int INF = 0x7FFFFFFF; int main() { // ifstream cin ("aaa.txt"); char val_a[maxn], val_b[maxn]; int f[maxn][maxn]; while (cin >> val_a + 1 >> val_b + 1) { val_a[0] = val_b[0] = '*'; int len_a = strlen (val_a), len_b = strlen (val_b); memset (f, 0, sizeof (f)); int i, j; for (i = 1; i < len_a; i++) { for (j = 1; j < len_b; j++) { if (val_a[i] == val_b[j]) f[i][j] = f[i-1][j-1] + 1; else f[i][j] = max (f[i-1][j], f[i][j-1]); } } int maxval = -INF; for (i = 1; i < len_b; i++) if (maxval < f[len_a-1][i]) maxval = f[len_a-1][i]; cout << maxval << endl; } return 0; } |
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