HDU 1159 Common Subsequence

　　　　　　　　　　　　　　　　　　　　　　　　 Common Subsequence

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

Submit Status Practice HDU 1159

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.

 

Input

abcfbc abfcab programming contest abcd mnp

 

Output

4 2 0

 

Sample Input

abcfbc abfcab programming contest abcd mnp

 

Sample Output

4 2 0

 

题目分析：

这个就是求两个字符串的最长子串。代码直接说吧

#include <cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<string.h>
using namespace std;

char str1[2000];
char str2[2000];
int dp[2000][2000];
int max(int a,int b )
{
    return (a>b?a:b);
}


int main()
{
    int i,j;
    int len1,len2;
    while(cin>>str1>>str2)
    {
        len1=strlen(str1);
        len2=strlen(str2);
        for(i=0;i<len1;i++)
        {
            dp[i][0]=0;
        }
        for(i=0;i<len2;i++)
        {
            dp[0][i]=0;
        }
        for(i=1;i<=len1;i++)
        {
            for(j=1;j<=len2;j++)
            {
                if(str1[i-1]==str2[j-1])
                {
                    dp[i][j]=dp[i-1][j-1]+1;
                }
                else
                {
                    dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
                }
            }
        }
        cout<<dp[len1][len2]<<endl;
    }
    return 0;
}