hdoj 1159最长公共子序列



/*Common Subsequence

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.
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*/

<span style="font-size:18px;">#include <stdio.h>  
#include <string.h>  
#define maxn 1000    
char str1[maxn], str2[maxn];  
int dp[maxn][maxn];    
int max(int a, int b)
{ 
return a > b ?
 a : b; 
}    
int gg()  
{  
    int m= 0;  
    for(int i = 1; str1[i]; ++i){  
        for(int j = 1; str2[j]; ++j){  
            if(str1[i] == str2[j]){  
                dp[i][j] = dp[i-1][j-1] + 1;                  
            }else dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);  
            if(dp[i][j]>m) m= dp[i][j];   
        }  
    }  
    return m;  
}  
  
int main()  
{  
    while(scanf("%s%s", str1 + 1, str2 + 1)== 2){  
        printf("%d
", gg());  
    }  
    return 0;  
}  
</span>