P4237 荒芜的海洋

我们先要保证奖赏尽量多，其次费用尽量小。应该可以看出，是一个最小费用最大流。如何建模？

考虑拆点。

每个岛屿分成一个入点一个出点。对于有雇佣兵的岛屿，源点向这个入点连容量为 $1$，费用为费用的边。对于怪兽的岛屿，连这个点的入点和出点，容量为 $+\infty$，费用为打怪兽的价格。对于桥梁，连出点和入点，容量为 $+\infty$，费用为经过的价格。对于有奖赏的岛，出点向汇点连边，容量为 $1$，费用为 $0$ 的边。

最小费用最大流即可，注意桥梁是双向边。

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

#define int long long

const int N = 5e5 + 5;

int n, m, a, b;
int S = 0, T = 0;
int e[N], h[N], c[N], cs[N], ne[N], idx;

void add(int u, int v, int w, int cc)
{
	//cout << "!!: " << u << " " << v << " " << w << " " << cc << "\n";
	e[idx] = v, c[idx] = w, ne[idx] = h[u], cs[idx] = cc, h[u] = idx++;
	e[idx] = u, c[idx] = 0, ne[idx] = h[v], cs[idx] = -cc, h[v] = idx++;
}

int dis[N], cur[N];
bool isin[N];

bool spfa()
{
	for (int i = 0; i <= T; i++)
	{
		isin[i] = 0;
		dis[i] = 5e9;
		cur[i] = -1;
	}
	queue<int> q;
	q.push(S);
	dis[S] = 0, cur[S] = h[S];
	while (q.size())
	{
		int u = q.front(); 
		q.pop();
		isin[u] = 0;
		for (int i = h[u]; ~i; i = ne[i])
		{
			int j = e[i];
			if (c[i] > 0 && dis[j] > dis[u] + cs[i])
			{
				//cout << j << "\n";
				cur[j] = h[j];
				dis[j] = dis[u] + cs[i];
				if (!isin[j])
				{
					q.push(j);
					isin[j] = 1;
				}
			}
		}
	}
	//cout << dis[T] << " " << dis[N - 1] << "\n";
	//system("pause");
	return (dis[T] != 5e9);
}

int mincost = 0;

int dfs(int u, int lim)
{
	if (u == T) return lim;
	isin[u] = 1;
	int sum = 0;
	for (int i = cur[u]; ~i && sum < lim; i = ne[i])
	{
		cur[u] = i;
		int j = e[i];
		if (dis[j] == dis[u] + cs[i] && c[i] > 0 && !isin[j])
		{
			int p = dfs(j, min(c[i], lim - sum));
			sum += p;
			c[i] -= p;
			c[i ^ 1] += p;
			mincost += p * cs[i];
		}
	}
	isin[u] = 0;
	return sum;
}

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

signed main()
{
	ios::sync_with_stdio(0), cin.tie(nullptr), cout.tie(nullptr);
	memset(h, -1, sizeof h);
	cin >> n >> m >> a >> b;
	for (int i = 1; i <= n; i++)
	{
		int x;
		cin >> x;
		add(i, i + n, INT_MAX, x);
	}
	T = 2 * n + 1;
	for (int i = 1; i <= m; i++)
	{
		int u, v, w;
		cin >> u >> v >> w;
		if (u == v) continue;
		add(u + n, v, INT_MAX, w);
		add(v + n, u, INT_MAX, w);
	}
	for (int i = 1; i <= a; i++)
	{
		int p, q;
		cin >> q >> p;
		add(S, p, 1, q);
	}
	int sum = 0LL;
	for (int i = 1; i <= b; i++)
	{
		int k, q;
		cin >> k >> q;
		sum += k;
		add(q + n, T, 1, 0);
	}
	int g = dinic();
	if (g == b)
	{
		cout << "Yes\n" << sum - mincost << "\n";
	}
	else
	{
		cout << "No\n" << g << "\n";
	}
	return 0;
}