hdu1159 - Common Subsequence (dp 最长公共子序列问题)

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22481    Accepted Submission(s): 9867

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
  
  

 

Source

Southeastern Europe 2003

 

 

              最长公共子序列问题是期末小学期前遗留下来的问题，今天上午又看了遍lcy的课件，恍然。曾经研究很长时间的辅助空间图，现在看意思就是判断a[i] 是否相等 b[j] 如果相等f[i][j] = f[i-1][j-1]+1 ; 如果不相等 f[i][j] = max(f[i][j-1], f[i][j-1]).

/********************************************
* 
*   acm:   hdu-1159
* 
*   title: Common Subsequence
* 
*   time:  2014.7.17 
* 
*********************************************/  

//考察dp 最长公共子序列问题：求公共子序列的最大和

/*
思路：
dp方程： f(i,j) = { 
                      f(i-1,j-1)+1             (a[i]==b[j])
					  max(f(i-1, j), f(i, j-1)) (a[i]!=b[j])
				  }
由于f(i,j)只和f(i-1,j-1), f(i-1,j)和f(i,j-1)有关,
而在计算f(i,j)时, 只要选择一个合适的顺序,
就可以保证这三项都已经计算出来了, 这样就可以计算出f(i,j). 
这样一直推到f(len(a),len(b))就得到所要求的解了.	
*/

#include <stdio.h>
#include <stdlib.h>

#define MAXSIZE 1000

typedef char String[MAXSIZE];

int f[MAXSIZE][MAXSIZE];

void dp(int i, int j, char a, char b)
{

    if (i == 0)  //边缘化
    {
        if (a == b)
        {
            f[i][j] = 1;
        }
        else
        {
            f[i][j] = f[i][j - 1];
        }
    }
    else if (j == 0)
    {
        if (a == b)
        {
            f[i][j] = 1;
        }
        else
        {
            f[i][j] = f[i - 1][j];
        }
    }
    else
    {
        if (a == b)
        {
            f[i][j] = f[i - 1][j - 1] + 1;
        }
        else
        {
            f[i][j] = f[i-1][j] > f[i][j - 1] ? f[i-1][j] : f[i][j - 1];
        }
    }
}

int main()
{
    String a, b;

    while (~scanf("%s%s", a, b))
    {
        int i;
        int j;

        //初始化
        if (a[0] == b[0])
        {
            f[0][0] = 1;
        }
        else
        {
            f[0][0] = 0;
        }

        for (i = 0; a[i] != '\0'; i++)
        {
            for (j = 0; b[j] != '\0'; j++)
            {
                dp(i, j, a[i], b[j]);
            }
        }

/*   //打印辅助空间变化图
       for (i = 0; a[i] != '\0'; i++)
        {
            for (j = 0; b[j] != '\0'; j++)
            {
                printf("%d ", f[i][j]);
            }
            printf("\n");
        }
*/
        printf("%d\n", f[i-1][j-1]);
    }

    return 0;
}