HDU 1159 Common Subsequence(dp LCS)

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HDU 1159 Common Subsequence(dp LCS)

来源: 作者: 时间:2016-01-23 10:51
Problem DescriptionA subsequence of a given sequence is the given sequence with some elements (possible none) left out Given a sequence X = another sequence Z = i

Problem Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = another sequence Z = is a subsequence of X if there exists a strictly increasing sequence of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = is a subsequence of X = 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

最长公共子序列




#include
#include
#include
using namespace std;

int max(int a,int b)
{
    return a>b? a:b;
}

char a[1001],b[1001];
int dp[1000][1000];

int main()
{
    while(scanf("%s%s",a,b)!=EOF)
    {
        int lena=strlen(a);
        int lenb=strlen(b);
        int i;

        for(i=0;i<=lena;i++)
            dp[i][0]=0;

        for(i=0;i<=lenb;i++)
            dp[0][i]=0;

        for(i=1;i<=lena;i++)
            for(int j=1;j<=lenb;j++)
            if(a[i-1]==b[j-1])
              dp[i][j]=dp[i-1][j-1]+1;
            else
              dp[i][j]=max(dp[i][j-1],dp[i-1][j]);//这个位置是矩形的左和上边

        printf("%d\n",dp[lena][lenb]);
    }
    return 0;
}


//下面样例是借鉴的,方便理解吧

/*

   p   r   o    g  r   a   m   n

   0   0   0   0   0   0   0   0   0

c  0   0   0   0   0   0   0   0   0

o  0   0   0   1   1   1   1   1   1

n  0   0   0   1   1   1   1   1   2

t  0   0   0   1   1   1   1   1   2

e  0   0   0   1   1   1   1   1   2

s  0   0   0   1   1   1   1   1   2

t  0   0   0   1   1   1   1   1   2

*/







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