POJ 3469 最小割

#include <cstdio>
#include <algorithm>
#include <vector>
#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAX_N = 2E4 + 10;
const int MAX_M = 2E5 + 10;
int N, M, A[MAX_N], B[MAX_N], b[MAX_M], a[MAX_M], w[MAX_M];
struct Edge
{
	int from, to, cap, flow;
	Edge(int from, int to, int cap, int flow): from(from), to(to), cap(cap), flow(flow) {}
};

struct Dinic
{
	int n, m, s, t;
	vector<Edge>edges;
	vector<int>G[MAX_N];
	bool vis[MAX_N];
	int d[MAX_N], cur[MAX_N];
	void add_edge(int from, int to, int cap)
	{
		edges.push_back(Edge(from, to, cap, 0));
		edges.push_back(Edge(to, from, 0, 0));
		m = edges.size();
		G[from].push_back(m - 2);
		G[to].push_back(m - 1);
	}
	bool BFS()
	{
		memset(vis, 0, sizeof(vis));
		queue<int>Q;
		Q.push(s);
		d[s] = 0;
		vis[s] = 1;
		while (!Q.empty())
		{
			int x = Q.front(); Q.pop();
			for (int i = 0; i < G[x].size(); i++)
			{
				Edge &e  = edges[G[x][i]];
				if (!vis[e.to] && e.cap > e.flow)
				{
					vis[e.to] = 1;
					d[e.to] = d[x] + 1;
					Q.push(e.to);
				}
			}
		}
		return vis[t];
	}
	int DFS(int x, int a)
	{
		if (x == t || a == 0) return a;
		int flow = 0, f;
		for (int &i = cur[x]; i < G[x].size(); i++)
		{
			Edge &e = edges[G[x][i]];
			if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0)
			{
				e.flow += f;
				edges[G[x][i] ^ 1].flow -= f;
				flow += f;
				a -= f;
				if (a == 0) break;
			}
		}
		return flow;
	}
	int max_flow(int s, int t)
	{
		this->s = s;
		this->t = t;
		int flow = 0;
		while (BFS())
		{
			memset(cur, 0, sizeof(cur));
			flow += DFS(s, INF);
		}
		return flow;
	}
};
Dinic D;
void solve()
{
	int s = N, t = s + 1;
	for (int i = 0; i < N; i++)
		D.add_edge(i, t, A[i]), D.add_edge(s, i, B[i]);
	for (int i = 0; i < M; i++)
		D.add_edge(a[i] - 1, b[i] - 1, w[i]), D.add_edge(b[i] - 1, a[i] - 1, w[i]);
	printf("%d\n", D.max_flow(s, t));
}
int main(int argc, char const *argv[])
{
	while (~scanf("%d%d", &N, &M) && N + M)
	{
		for (int i = 0; i < N; i++)
			scanf("%d%d", &A[i], &B[i]);
		for (int i = 0; i < M; i++)
			scanf("%d%d%d", &a[i], &b[i], &w[i]);
		solve();
	}
	return 0;
}

这道题目白书上有题解，可惜他的算法过不了时限。。。。。

需要用刘汝佳的算法，Dinic和ISAP都可以。