Poj 1458 Common Subsequence 최장 비연속 시퀀스

제목 설명

가장 긴 동일한 서열, 비연속, abc와ac의 가장 긴 동일한 서열을 2
관련 질문, 스탬프:
최장 공통 서열 길이 최장 상승 서열 길이

아이디어:

dp[i][j]표시: a열에서 전 i개 원소와 b에서 전 j개 원소 중 가장 길고 같은 서열 길이는 두 가지로 나뉜다. 1)a[i]==b[j]는 dp[i][j]=dp[i-1][j-1]+1 위에 앉는 +12)a[i]=!b[j]는 dp[i][j]=max(dp[i-1][j], dp[i][j-1]) 정위와 정좌가 가장 크다.

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

/****************************** > File Name: poj_1458.cpp > Author: dulun > Mail: [email protected] > Created Time: 2016 04 19      16 48 16  *************************/

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#define LL long long
using namespace std;

const int N = 666;
char a[N];
char b[N];
int dp[N][N];

int main()
{
    while(~scanf("%s%s", a+1, b+1))
    {
        int ans = 0;
        int len_a = strlen(a+1);
        int len_b = strlen(b+1);
        memset(dp, 0, sizeof(dp));

        for(int i = 1; i <= len_a; i++)
        {
            for(int j = 1; j <= len_b; j++)
            {
                if(a[i] != b[j] )
                {
                    dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
                }
                else{
                    dp[i][j] = dp[i-1][j-1]+1;
                }
                ans = max(ans, dp[i][j]);
            }
        }
        printf("%d
", ans);
    }
    return 0;
}