「NOI2006」最大获利

「NOI2006」最大获利

传送门

最小割。

对于每一组用户群 \(A_i, B_i, C_i\) ，连边 $S \to A_i, S \to B_i, $ 容量为成本，还有 \(i \to T\) ，容量为收益 \(C_i\)，\(A_i, B_i\) 都向 \(i\) 连边，容量为 \(inf\) ，割掉与 \(S\) 的连边表示支付成本，割掉与 \(T\) 的连边表示放弃收益。

参考代码：

#include <algorithm>
#include <cstring>
#include <cstdio>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
using namespace std;
template < class T > inline void read(T& s) {
	s = 0; int f = 0; char c = getchar();
	while ('0' > c || c > '9') f |= c == '-', c = getchar();
	while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
	s = f ? -s : s;
}

const int _ = 55010, __ = 155010, INF = 2147483647;

int tot = 1, head[_];
struct Edge { int ver, cap, nxt; } edge[__ << 1];
inline void Add_edge(int u, int v, int d) { edge[++tot] = (Edge) { v, d, head[u] }, head[u] = tot ; }
inline void link(int u, int v, int d) { Add_edge(u, v, d), Add_edge(v, u, 0); }

int n, m, s, t, dep[_], cur[_], Q[_];

inline int bfs() {
	int hd = 0, tl = 0;
	memset(dep, 0, sizeof (int) * (t - s + 1));
	Q[++tl] = s, dep[s] = 1;
	while (hd < tl) {
		int u = Q[++hd];
		for (rg int i = head[u]; i; i = edge[i].nxt) {
			int v = edge[i].ver;
			if (dep[v] == 0 && edge[i].cap > 0)
				dep[v] = dep[u] + 1, Q[++tl] = v;
		}
	}
	return dep[t] > 0;
}

inline int dfs(int u, int flow) {
	if (u == t) return flow;
	for (rg int& i = cur[u]; i; i = edge[i].nxt) {
		int v = edge[i].ver;
		if (dep[v] == dep[u] + 1 && edge[i].cap > 0) {
			int res = dfs(v, min(flow, edge[i].cap));
			if (res) { edge[i].cap -= res, edge[i ^ 1].cap += res; return res; }
		}
	}
	return 0;
}

inline int Dinic() {
	int res = 0;
	while (bfs()) {
		for (rg int i = s; i <= t; ++i) cur[i] = head[i];
		while (int d = dfs(s, INF)) res += d;
	}
	return res;
}

int main() {
#ifndef ONLINE_JUDGE
	file("cpp");
#endif
	int sum = 0;
	read(n), read(m), s = 0, t = n + m + 1;
	for (rg int x, i = 1; i <= n; ++i) read(x), link(s, i, x);
	for (rg int x, y, z, i = 1; i <= m; ++i)
		read(x), read(y), read(z), link(x, i + n, INF), link(y, i + n, INF), link(i + n, t, z), sum += z;
	printf("%d\n", sum - Dinic());
	return 0;
}