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HDU 1159 Common Subsequence 最大公共子序列

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HDU 1159 Common Subsequence 最大公共子序列

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

這道題就是求最大公共子序列的長度。

不知道怎麼解釋。

只好打了個草圖。

#include 
#include 
#define max(a,b) a>b?a:b
char s[520];
char s1[520];
int lcs[520][520];
int LCS(int l,int l1)
{
    int i,j;  //將兩列字符竄變成i行,j列。lsc數組代表每一個位置的最大公共子序列的長度。
    for(i=1;i<=l;i++)   
        for(j=1;j<=l1;j++)  //將s[i-1]分別和s1的每一個元素做比較
        {
            if(s[i-1]==s1[j-1])  //碰到相等的。
                lcs[i][j]=lcs[i-1][j-1]+1;//圖中加一的情況
            else
                lcs[i][j]=max(lcs[i-1][j],lcs[i][j-1]);//碰到不相等的，則取它的上方和左方的那個的最大值，圖中都為一。以此累加
        } 
        return lcs[l][l1];  //到達最後的狀態必然是最大的長度,圖中的長度為2，最大公共子序列可以是ca或者ab。
}
int main()
{
    while(scanf("%s%s",s,s1)!=EOF)
    {
        int l=strlen(s);
        int l1=strlen(s1);
        memset(lcs,0,sizeof(lcs));
        printf("%d\n",LCS(l,l1));
    }
    return 0;
}