POJ 1458 Common Subsequence

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1458

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, x _ij = 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.

Input

The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.

Output

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

程序分析：此题也是求最长公共子序列。和上一篇很相似，基本可以套用模板同样是DP。在处理时可以选择字符的读取从第一个位置开始，或者把 i 号字符的状态存储到i+1号位置去，这样就从1号开始处理了，判定是就是 s1[i-1] == s1[j-1]。

程序代码：

#include <cstring>
#include <cstdlib>
#include <cstdio> 
#include<iostream>
#define Max( a, b ) (a) > (b) ? (a) : (b) 
using namespace std;

char s1[1005], s2[1005]; 
int dp[1005][1005];
int main()
{
    int len1, len2;   
    while( scanf( "%s %s", s1+1, s2+1 ) != EOF )
    {        
        memset( dp, 0, sizeof(dp) );  
        len1 = strlen( s1+1 ), len2 = strlen( s2+1 );    
        for( int i = 1; i <= len1; ++i )    
        {      
            for( int j = 1; j <= len2; ++j )        
            {               if( s1[i] == s2[j] )          
            {                   dp[i][j] = dp[i-1][j-1] + 1;    
            }                 else                
            {               
                
                dp[i][j] = Max ( dp[i-1][j], dp[i][j-1] );     
            }
            }
        }  
        printf( "%d
", dp[len1][len2] );  
    }   
    return 0; 
}