hdu1159Common Subsequence（动态规划之最长公共子序列）

Common Subsequence

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 5   Accepted Submission(s) : 3

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

   
   
   
   
   
   
   
   

    
    
    
    题意：
   
   
   
   
   
   
   
   

    
    
    
    最长公共子序列；
   
   
   
   
   
   
   
   

    
    
    
    #include<cstring>
#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
char a[10000],b[10000];
int dp[1000][1000];
int main()
{
    while(~scanf("%s %s",a,b))
    {


        memset(dp,0,sizeof(0));
       int i,j;
       int m,n;
       m=strlen(a);
       n=strlen(b);
       for(i=1;i<=m;i++)
       {
           for(j=1;j<=n;j++)
           {
               if(a[i-1]==b[j-1])
               {
                   dp[i][j]=dp[i-1][j-1]+1;//判断是否相等。
               }
               else
                dp[i][j]=max(dp[i-1][j],dp[i][j-1]);//判断是从哪个下来的
           }
       }
       printf("%d\n",dp[m][n]);


    }
    return 0;
}