hdu1159

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10900    Accepted Submission(s): 4464

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

 

这个是一个求最长公共子序列的一个问题。。：

代码如下：

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
char str1[1000];
char str2[1000];
int map[1000][1000];
void cal()
{
    int m=strlen(str1+1);
    int n=strlen(str2+1);
    for(int i=0;i<=m;i++)
    {
        map[i][0]=0;
    }
    for(int i=0;i<=n;i++)
    {
        map[0][i]=0;
    }
    for(int i=1;i<=m;i++)
    {
        for(int j=1;j<=n;j++)
        {
            if(str1[i]==str2[j])
            {
                map[i][j]=map[i-1][j-1]+1;
            }
            else
            {
                map[i][j]=max(map[i-1][j],map[i][j-1]);
            }
        }
    }
    cout<<map[m][n]<<endl;
}
int main()
{
    while(scanf("%s%s",str1+1,str2+1)==2)
    {
        cal();
    }
    return 0;
}