hdu 1159 Common Subsequence

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本文介绍了一种解决字符串最长公共子序列问题的方法，通过动态规划技术找到两个给定序列之间的最大公共子序列长度。示例输入输出展示了如何应用此算法，并提供了几个相似问题供进一步学习。

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//开始DP之旅,先来道简单的

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

Source

Southeastern Europe 2003

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/********************

求字符串 A 和字符串 B 最长公共子序列(序列可以不连着)

用一个二维数组 num[][] 存当前最大的子序列长度

第一组测试数据的num[][] 值

所以可以得到递推方程:

A[i] = B[j] 时,num[i][j] = num[i-1][j-1]+1;

A[i] = B[j] 时,num[i][j] = max_(num[i-1][j] , num[i][j-1]);

******************/

Code:

#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
#define N 1000
int num[N][N];
int max_(int a,int b)
{
    return a>b?a:b;
}
int main()
{
    char A[N],B[N];
    int i,j,a,b;
    while(scanf("%s%s",A,B)!=EOF)
    {
        memset(num,0,sizeof(num));
        a = strlen(A);
        b = strlen(B);
        for(i = 1;i<=a;i++)
        {
            for(j = 1;j<=b;j++)
            {
                if(A[i-1]==B[j-1])
                    num[i][j] = num[i-1][j-1]+1;
                else
                    num[i][j] = max_(num[i-1][j],num[i][j-1]);
            }
        }
        printf("%d\n",num[a][b]);
    }
    return 0;
}