Codeforces Round 977 (Div. 2）E1 Digital Village (Easy Version)（Floyd，贪心）

题目链接

Codeforces Round 977 (Div. 2）E1 Digital Village (Easy Version)

思路

首先，我们注意到 n n n的最大值只有 400 400 400。

因此，我们可以先用 F l o y d Floyd Floyd算法预处理出任意两座城市之间的最大延迟时间。

之后，我们通过在线操作，每次贪心地选出最优的一个城市，并不断更新答案。

即，我们先选出 k = 1 k=1 k=1时的最优解，之后从剩下的点里面挑出一个能够使 k = 2 k=2 k=2时最优的点，以此类推。

代码

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 4e2 + 5;
const int mod = 998244353;
const int inf = 0x3f3f3f3f3f3f3f3f;
int n, m, p;
void solve()
{
	cin >> n >> m >> p;
	vector<int>city(p);
	for (int &val : city)
	{
		cin >> val;
	}
	vector<vector<int>>dist(n + 1, vector<int>(n + 1, inf));
	for (int i = 1, u, v, w; i <= m; i++)
	{
		cin >> u >> v >> w;
		dist[u][v] = min(dist[u][v], w);
		dist[v][u] = min(dist[v][u], w);
	}

	for (int i = 1; i <= n; i++)
		dist[i][i] = 0;

	for (int k = 1; k <= n; k++)
	{
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				dist[i][j] = min(dist[i][j], max(dist[i][k], dist[k][j]));
			}
		}
	}


	vector<int>ans(n + 1);

	int sum = inf, idx = -1;
	for (auto u : city)
	{
		int res = 0;
		for (auto v : city)
		{
			res += dist[u][v];
		}
		if (sum >= res)
		{
			idx = u;
			sum = res;
			for (auto v : city)
			{
				ans[v] = dist[u][v];
			}
		}
	}

	set<int>st;
	st.insert(idx);

	cout << sum << " ";

	for (int i = 2; i <= n; i++)
	{
		if (sum == 0)
		{
			cout << sum << " ";
			continue;
		}
		int maxx = 0;
		idx = -1;
		for (auto u : city)
		{
			if (st.count(u)) continue;
			int res = 0;
			for (auto v : city)
			{
				if (st.count(v)) continue;
				if (ans[v] > dist[u][v])
				{
					res += ans[v] - dist[u][v];
				}
			}
			if (maxx <= res)
			{
				maxx = res;
				idx = u;
			}
		};
		st.insert(idx);
		sum -= maxx;
		for (auto v : city)
		{
			if (st.count(v)) continue;
			ans[v] = min(ans[v], dist[idx][v]);
		}
		cout << sum << " ";
	}
	cout << endl;
}
signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	int test = 1;
	cin >> test;
	for (int i = 1; i <= test; i++)
	{
		solve();
	}
	return 0;
}