目錄

Common Subsequence

題目解釋：

解題思路：

ac程式碼：

Common Subsequence

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 26   Accepted Submission(s) : 15

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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

題目解釋：

求兩個字串的最長公共子序列的長度（子序列可以不連續）

解題思路：

狀態轉移方程：

if (a[i]==b[j])

dp[i][j]=dp[i-1][j-1]+1;

else

dp[i][j]=max(dp[i-1][j],dp[i][j-1]);

ac程式碼：

注意字串的讀取語句

#include <iostream>
 #include <cstring>
#define maxn 1005
using namespace std;
int main()
{
    int dp[maxn][maxn];//dp[i][j]表示a的第i個字元和b的第j個字元的LCS
    char a[maxn],b[maxn];
    while(cin>>a+1>>b+1)
    {
        int lena=strlen(a+1),lenb=strlen(b+1);
        int i,j;//i指向a，j指向b
        for(i=0;i<=lena;i++)//i從0開始，因為下面有i-1
            dp[i][0]=0;
        for(j=0;j<=lenb;j++)
            dp[0][j]=0;
        for(i=1;i<=lena;i++)
        {
           for(j=1;j<=lenb;j++)
          {
            if(a[i]==b[j])
                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[lena][lenb]);
    }
    return 0;
}