|Time Limit: 1000MS||Memory Limit: 10000K|
|Total Submissions: 22340||Accepted: 8524|
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 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.
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.
abcfbc abfcab programming contest abcd mnp
4 2 0
Southeastern Europe 2003
Basically, it's a straight forward dynamic programming problem----longest common subsequence.
It's pretty much like POJ1159, I used the same subfunction to get the answer.
Source Code://Tue Apr 13 15:46:54 CDT 2010
using namespace std;
int LongestIncreasingCommonSubsequence(string A, string B)
int N = A.size();
int M = B.size();
vector<vector<int> > v(N + 1, vector<int>(M + 1, 0));
for (int i = 0; i <= N; i++)
for (int j = 0; j <= M; j++)
if (i == 0 || j == 0)
v[i][j] = 0;
else if(A[i-1] == B[j-1])
v[i][j] = v[i-1][j-1] + 1;
v[i][j] = max(v[i-1][j], v[i][j-1]);
int main(int argc, const char* argv)
freopen("input.in", "r", stdin);
freopen("output.out", "w", stdout);
string str1, str2;
while(cin >> str1 >> str2)
cout << LongestIncreasingCommonSubsequence(str1, str2) << endl;