【水:最长公共子序列】【HDU1159】【Common Subsequence】


Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24919Accepted Submission(s): 11035


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
题目没看完 看了一下样例 估计是最长公共子序列。。
虽然水题 还是记不牢 写一遍

#include <cstdio>  
#include <cstdlib>  
#include <cmath>  
#include <cstring>  
#include <ctime>  
#include <algorithm>  
#include <iostream>
#include <sstream>
#include <string>
#define oo 0x13131313   
using namespace std;
char A[1001],B[1001];
int F[1000][1000];
void solve()
{
		memset(F,0,sizeof(F));
		int lena=strlen(A);
		int lenb=strlen(B);
		F[0][0]=0;
		F[0][1]=0;
		F[1][0]=0;
		for(int i=1;i<=lena;i++)
		 for(int j=1;j<=lenb;j++)
	{
		 if(A[i-1]==B[j-1])
		 F[i][j]=F[i-1][j-1]+1;
		 else F[i][j]=max(F[i][j-1],F[i-1][j]);
	}
		printf("%d\n",F[lena][lenb]);
}
int main()
{
	while(scanf("%s %s",A,B)!=EOF)
	{
		solve();
	}
	return 0;
}
  


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