NETADMIN - Smart Network Administrator

有趣题。

容易发现答案可以二分，重点在于怎么写 check。

我们假设现在二分的是 $x$，我们要判断若对于每条边，只能经过 $x$ 次，$k$ 个点是否可以全部到达 $1$。容易发现这个东西可以转化为网络流，从源点 $S$ 向 $k$ 个特殊点连容量为 $1$ 的边，原图上每条边的容量为 $x$。由题知，汇点 $T=1$，我们只需要判断 $S \rightarrow T$ 的最大流是否为 $k$ 即可。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <queue>
#include <climits>
using namespace std;

const int N = 2e5 + 5;

int t, n, m, k;
int e[N], h[N], ne[N], c[N], idx;
int S = 0, T = 1;
int d[N], cur[N];
int a[N];
int u[N], v[N];

inline void add(int x, int y, int z)
{
	e[idx] = y, ne[idx] = h[x], c[idx] = z, h[x] = idx++;
}

inline bool bfs()
{
	queue<int> q;
	q.push(S);
	for (int i = 0; i <= n; i++) d[i] = -1, cur[i] = -1;
	d[S] = 0, cur[S] = h[S];
	while (q.size())
	{
		int u = q.front();
		q.pop();
		for (int i = h[u]; ~i; i = ne[i])
		{
			int j = e[i];
			if (d[j] == -1 && c[i] > 0)
			{
				d[j] = d[u] + 1;
				cur[j] = h[j];
				if (j == T) return 1;
				q.push(j);
			}
		}
	}
	return 0;
}

inline int dfs(int u, int lim)
{
	if (u == T) return lim;
	int res = 0;
	for (int i = cur[u]; ~i && res < lim; i = ne[i])
	{
		cur[u] = i;
		int j = e[i];
		if (d[j] == d[u] + 1 && c[i] > 0)
		{
			int p = dfs(j, min(lim - res, c[i]));
			if (p == 0) d[j] = -1;
			res += p;
			c[i] -= p;
			c[i ^ 1] += p;
		}
	}
	return res;
}

inline int dinic()
{
	int res = 0, p;
	while (bfs())
	{
		while (p = dfs(S, INT_MAX)) res += p;
	}
	return res;
}

inline bool check(int g)
{
	for (int i = 0; i < N; i++)
	{
		e[i] = h[i] = ne[i] = c[i] = 0;
		h[i] = -1;
	}
	idx = 0;
	for (int i = 1; i <= k; i++)
	{
		add(S, a[i], 1);
		add(a[i], S, 0);
	}
	for (int i = 1; i <= m; i++)
	{
		add(u[i], v[i], g);
		add(v[i], u[i], g);
	}
	int gg = dinic();
	//printf("!: %d %d\n", g, gg);
	return gg == k;
}

int main()
{
	memset(h, -1, sizeof h);
	scanf("%d", &t);
	while (t--)
	{
		idx = 0;
		scanf("%d%d%d", &n, &m, &k);
		for (int i = 1; i <= k; i++) scanf("%d", &a[i]);
		for (int i = 1; i <= m; i++)
		{
			scanf("%d%d", &u[i], &v[i]);
		}
		int l = 0, r = k + 1, ans = -1;
		while (l <= r)
		{
			int mid = l + r >> 1;
			if (check(mid)) ans = mid, r = mid - 1;
			else l = mid + 1;
		}
		printf("%d\n", ans);
	}
	return 0;
}