【HDU】1159 - Common Subsequence（LCS）

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Common Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 34330 Accepted Submission(s): 15666

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.

Sample Input

   abcfbc abfcab
programming contest 
abcd mnp

Sample Output

   4
2
0

Source

Southeastern Europe 2003

LCS的模板题。

代码如下：

#include <stdio.h>
#include <cstring>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define LL long long
int dp[1000][1000];
int main()
{
	char a[1000],b[1000];
	int l1,l2;
	while (~scanf ("%s %s",a,b))
	{
		l1 = strlen(a);
		l2 = strlen(b);
		for (int i = 0 ; i <= max(l1,l2) ; i++)
			dp[i][0] = dp[0][i] = 0;
		for (int i = 1 ; i <= l1 ; i++)
		{
			for (int j = 1 ; j <= l2 ; j++)
			{
				if (a[i-1] == b[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[l1][l2]);
	}
	return 0;
}