P6768 [USACO05MAR]Ombrophobic Bovines 发抖的牛

可以发现，牛如果想要去其他地点的雨棚，那么一定走最短路。

所以可以先 $O(n^3)$ 处理最短路，然后二分答案，将所有最短路 $\leq$ 当前二分的时间的边加入。

然后网络流判断可行性，即超级源点向每一块田地连一条容量为牛的个数的边，然后把田地和雨棚拆点，雨棚向超级汇点连容量为雨棚最多能放牛的个数的边。最后两点判断一次最短路，看是否可以加边。

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

#define int long long

const int N = 5e6, INF = INT_MAX;

int dis[205][205], cow[N], rain[N], sum;

int n, p, T, S;
int e[N], h[N], c[N], ne[N], idx;

int cur[N], d[N];

void add(int x, int y, int z)
{
	e[idx] = y, ne[idx] = h[x], c[idx] = z, h[x] = idx++;
	e[idx] = x, ne[idx] = h[y], c[idx] = 0, h[y] = idx++;
}

void solve(int maxn)
{
	for (int i = 0; i <= T; i++) h[i] = -1;
	for (int i = 0; i <= idx; i++)
	{
		c[i] = ne[i] = e[i] = 0;
	}
	for (int i = 1; i <= n; i++)
	{
		add(S, i, cow[i]);
	}
	for (int i = 1; i <= n; i++)
	{
		add(i + n, T, rain[i]);
	}
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= n; j++)
		{
			if (dis[i][j] <= maxn)
			{
				add(i, j + n, INF);
				//add(j, i, INF);
				//printf("%lld %lld\n", i, j);
			}
		}
	}
}

bool bfs()
{
	for (int i = 0; i <= T; i++) d[i] = -1;
	queue<int> q;
	q.push(S);
	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;
}

int dfs(int u, int flow)
{
	if (u == T) return flow;
	int sum = 0;
	for (int i = cur[u]; ~i && sum < flow; i = ne[i])
	{
		int j = e[i];
		cur[u] = i;
		if (d[j] == d[u] + 1 && c[i] > 0)
		{
			int f = dfs(j, min(c[i], flow - sum));
			if (!f) d[j] = -1;
			else
			{
				sum += f;
				c[i] -= f;
				c[i ^ 1] += f;
			}
		}
	}
	return sum;
}

int dinic()
{
	int sum = 0, f;
	while (bfs())
	{
		while (f = dfs(S, INF))
		{
			sum += f;
		}
	}
	return sum;
}

bool check(int x)
{
	solve(x);
	return dinic() >= sum;
}

signed main()
{
	memset(dis, 0x3f, sizeof dis);
	scanf("%lld%lld", &n, &p);
	T = 2 * n + 1;
	for (int i = 1; i <= n; i++) scanf("%lld%lld", &cow[i], &rain[i]), dis[i][i] = 0, sum += cow[i];
	for (int i = 1; i <= p; i++)
	{
		int u, v, w;
		scanf("%lld%lld%lld", &u, &v, &w);
		dis[u][v] = min(dis[u][v], w);
		dis[v][u] = min(dis[v][u], w);
	}
	for (int k = 1; k <= n; k++)
	{
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
			}
		}
	}
	bool f = 0;
	int l = 0, r = INF * 100;
	while (l <= r)
	{
		int mid = l + r >> 1;
		if (check(mid)) r = mid - 1, f = 1;
		else l = mid + 1;
	}
	printf("%lld\n", f ? r + 1 : -1);
	return 0;
}