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Codeforces #264 (Div. 2) D. Gargari and Permutations

2024-07-20 12:01:39   222人阅读

Gargari got bored to play with the bishops and now, after solving the problem about them, he is trying to do math homework. In a math book he have foundk permutations. Each of them consists of numbers1,?2,?...,?n in some order. Now he should find the length of the longest common subsequence of these permutations. Can you help Gargari?

You can read about longest common subsequence there: https://en.wikipedia.org/wiki/Longest_common_subsequence_problem

Input

The first line contains two integers n andk (1?≤?n?≤?1000; 2?≤?k?≤?5). Each of the nextk lines contains integers 1,?2,?...,?n in some order — description of the current permutation.

Output

Print the length of the longest common subsequence.

Sample test(s)
Input
4 3
1 4 2 3
4 1 2 3
1 2 4 3
Output
3
Note

The answer for the first test sample is subsequence [1, 2, 3].

题意：求k个长度为n的最长公共子序列

思路：保存每个数在各自串的位置，那么如果一个以j为结束的最长公共子序列成立的情况是，对于每个串的i都在j的前面，那么就有dp[j] = max(dp[j], dp[i]+1)

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 1010;

int n, k;
int a[maxn][maxn], b[maxn][maxn], dp[maxn];

int check(int x, int y) {
	for (int i = 2; i <= k; i++)
		if (b[i][x] > b[i][y])
			return 0;
	return 1;
}

int main() {
	scanf("%d%d", &n, &k);
	for (int i = 1; i <= k; i++)
		for (int j = 1; j <= n; j++) {
			scanf("%d", &a[i][j]);
			b[i][a[i][j]] = j;
		}

	for (int i = 1; i <= n; i++)
		dp[i] = 1;

	int ans = 0;
	for (int i = 1; i <= n; i++) {
		for (int j = i+1; j <= n; j++) {
			if (check(a[1][i], a[1][j]))
				dp[j] = max(dp[i]+1, dp[j]);
		}
	}

	for (int i = 1; i <= n; i++)
		ans = max(ans, dp[i]);
	printf("%d\n", ans);
	return 0;
}


Codeforces #264 (Div. 2) D. Gargari and Permutations


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