HDU 1159 Common Subsequence 公共子序列 DP 水题重温

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1159

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18201    Accepted Submission(s): 7697

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

 

解题思路：

开一个二维数组f[N][M]记录第一个数组取前N个数和第二个数组取前M个数的时候公共子序列的最大长度，最后输出f[l1][l2]即为最后结果。

 

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
char keng1[2000],keng2[2000];
int con[2000][2000];
int getmax(int a,int b)
{if(a>b)return a;else return b;}
int main()
{
    int i,j,l1,l2,Max;
    keng1[0]=keng2[0]='#';
    while(scanf("%s%s",keng1+1,keng2+1)!=EOF)
    {
        Max=0;
        l1=strlen(keng1);
        l2=strlen(keng2);
        memset(con,0,sizeof(con));
        for(i=1;i<l1;i++)
        {
            for(j=1;j<l2;j++)
            {
                if(keng1[i]==keng2[j])
                    con[i][j]=getmax(con[i][j],con[i-1][j-1]+1);
                else
                con[i][j]=getmax(con[i-1][j],con[i][j-1]);
            }
        }
        printf("%d
",con[l1-1][l2-1]);
    }
    return 0;
}