解题报告 HDU1159 Common Subsequence

Common Subsequence

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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

 

 

分析：

　　设c[i][j]为字符串a的第i个字符与字符串b的第j个字符为止的最长公共子序列长度，那么有两种情况：

1. 当a[i] == b[j]时，c[i][j]应该是前一个状态的最长公共子序列长度 + 1，因为a[i]与b[j]匹配，a[i]与b[j]必然不能已经匹配过，否则就是同一个字母匹配了多次，这必然是非法的，因此上一个状态应是c[i - 1][j - 1]，即c[i][j] = c[i - 1][j - 1] + 1；
2. 当a[i] != b[j]时，上一个状态可能是c[i - 1][j]或c[i][j - 1]，而既然要找最长公共子序列，自然是找最大的一个，即c[i][j] = max(c[i - 1][j], c[i][j - 1])。

AC代码：

#include<iostream>
#include<string>
using namespace std;
int dp[1005][1005];
int main()
{
    string a,b;
    while(cin>>a>>b)
    {
        int alen=a.length();
        int blen=b.length();
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=alen;i++)
        {
            for(int j=1;j<=blen;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]);
            }
        }
        for(int i=1;i<=alen;i++)
        {
            for(int j=1;j<=blen;j++)
                cout<<dp[i][j]<<" ";
            cout<<endl;
        }
        cout<<dp[alen][blen]<<endl;
    }
    return 0;
}