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Common Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 49216 Accepted Submission(s): 22664
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

代碼：

#include<iostream>
#include<string.h>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int  maxnum = 1000 + 5;
 
int dp[maxnum][maxnum];
#define max(x,y){x>y?x:y}
void DP_LCS(char str1[], char str2[])
{
    memset(dp, 0, sizeof(dp));
    int i, j;
    for (i = 0; i < strlen(str1); i++)
    {
        for (j = 0; j < strlen(str2); j++)
        {
            if ( strlen(str1)==0 || strlen(str2) == 0)//邊界情況：如果有個字符串長度為0
            {
                dp[i+1][j+1] = 0;//公共子序列為0
            }
            if (str1[i] == str2[j])//第一種情況：a[i]==b[j]  A的前i個,B的前j個;
            {
                dp[i+1][j+1] = dp[i][j] + 1;//直接加1
            }
            else//第二、三種情況 dp[i][j]=dp[i-1][j]||dp[i][j]=dp[i][j-1]  A的前i-1個,B的前j個;A的前i個,B的前j-1個;
            {
                dp[i+1][j+1] = max(dp[i][j+1], dp[i+1][j]);
            }
        }
    }
    cout<< dp[strlen(str1)][strlen(str2)]<<endl;
}
int main()
{
    char  str1[maxnum], str2[maxnum];
    int N;
    while (cin >> str1 >> str2)
    {
        DP_LCS(str1, str2);
    }
    return 0;
}

hdu1159-Common Subsequence（DP：最長公共子序列LCS）