2020寒假集训排位赛 Milk Factory 题解（floyd）

专题链接：https://codeforces.com/group/5yyKg9gx7m/contest/269908/problem/F

思路

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最基础的floyd用法，能通即为1，不能通即为0，用floyd处理一遍即可

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<string>
#include<map>
#include<cstring>
#include<climits>
#include<vector>
#include<queue>
#include<stack>
#define lowbit(i) ((i)&(-(i)))
#define ll long long
using namespace std;
int N;
int m[101][101];
int main()
{
	cin >> N;
	for (int i = 1; i <= N-1; i++)
	{
		int x, y;
		scanf("%d%d", &x, &y);
		m[x][y] = 1;
	}
	for (int k = 1; k <= N; k++)
	{
		for (int i = 1; i <= N; i++)
		{
			for (int j = 1; j <= N; j++)
				if (i != j && j != k && k != i&& m[i][k]==1&& m[k][j]==1)
					m[i][j] = 1;
		}
	}
	int ans=0;
	for (int j = 1; j <= N; j++)
	{
		bool pan = 1;
		for (int i = 1; i <= N; i++)
		{
			if (i == j)
				continue;
			if (m[i][j] == 0)
			{
				pan = 0;
				break;
			}
		}
		if (pan)
		{
			ans = j;
          break;
		}
	}
	if (ans)
		printf("%d\n", ans);
	else
		printf("-1\n");
	return 0;
}