POJ 1458 Common Subsequence

Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 33069		Accepted: 12966

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
解题方法：最长公共子序列，直接套用模版。

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;

int main()
{
    char str[300];
    char substr[300];
    int dp[300][300];
    while(scanf("%s %s", str, substr) != EOF)
    {
        int nLen = strlen(str);
        int nLen1 = strlen(substr);
        for (int i = 0; i <= nLen; i++)
        {
            dp[i][0] = 0;
        }
        for (int i = 0; i <= nLen1; i++)
        {
            dp[0][i] = 0;
        }
        for (int i = 1; i <= nLen; i++)
        {
            for (int j = 1; j <= nLen1; j++)
            {
                if (str[i - 1] == substr[j - 1])
                {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                }
                else
                {
                    dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]);
                }
            }
        }
        printf("%d\n", dp[nLen][nLen1]);
    }
    return 0;
}