zoj 3362 Beer Problem 最小费用流

       别人都说是赤裸裸的最小费用流，自己想了很久，都想不到如何做，最后看了别人解题报告才懂。

       关键还是在构图和转化。源点是1，添加一个汇点，汇点到各个城市的边的容量是各个城市卖啤酒的钱的相反数。通过spfa求cost的最短路径，若最短路径的cost小于0，说明这条路径是赚钱的，否则是亏本的，算法停止。

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

const int N = 105;
const int M = 2005;
const int INF = 1000000000;

struct edge
{
	int v;
	int cap;
	int cost;
	int next;
};

edge E[M*4+N*2];
int head[N];
int cost[N];
int pre_v[N], pre_e[N];
bool inQ[N];

int n, m;
int cnt;

void add_edge(int u, int v, int cap, int cost)
{
	E[cnt].v = v;
	E[cnt].cap = cap;
	E[cnt].cost = cost;
	E[cnt].next = head[u];
	head[u] = cnt++;

	E[cnt].v = u;
	E[cnt].cap = 0;
	E[cnt].cost = -cost;
	E[cnt].next = head[v];
	head[v] = cnt++;
}

int mcmf(int s, int t, int n)
{
	queue<int> Q;
	int u;
	int mincost = 0;
	while (1)
	{
		memset(inQ, false, sizeof(inQ));
		//memset(pre_e, -1, sizeof(pre_e));
		memset(pre_v, -1, sizeof(pre_v));
		Q.push(s);
		for (int i = 0; i < n; i++)
			cost[i] = INF;
		cost[s] = 0;
		inQ[s] = true;

		//spfa
		//根据每条边的cost求最短路
		while (!Q.empty())
		{
			u = Q.front();
			Q.pop();
			inQ[u] = false;

			for (int i = head[u]; i != -1; i = E[i].next)
			{
				int v = E[i].v;
				if (E[i].cap && cost[u] + E[i].cost < cost[v])
				{
					cost[v] = cost[u] + E[i].cost;
					if (!inQ[v])
					{
						Q.push(v);
						inQ[v] = true;
					}
					
					pre_v[v] = u; pre_e[v] = i;
				}
			}
		}

		//当cost[t]大于0时，说明当前的买卖时亏本的，退出
		if (cost[t] >= 0)
			break;

		int f = INF;
		for (int i = t; i != s; i = pre_v[i])
		{
			int v = pre_e[i];
			f = min(f, E[v].cap);
		}

		mincost += f*cost[t];

		for (int i = t; i != s; i = pre_v[i])
		{
			int u = pre_e[i];
			E[u].cap -= f;
			E[u^1].cap += f;
		}

	}

	return -mincost;
}

int main()
{
	int u, v, cap, w;
	while (cin >> n >> m)
	{
		if (!n && !m)
			break;
		
		cnt = 0;
		memset(head, -1, sizeof(head));
		for (int i = 2; i <= n; i++)
		{
			scanf("%d", &w);
			add_edge(i, n+1, INF, -w);
		}

		for (int i = 0; i < m; i++)
		{
			scanf("%d%d%d%d", &u, &v, &cap, &w);
			add_edge(u, v, cap, w);
			add_edge(v, u, cap, w);
		}

		cout << mcmf(1, n+1, n+2) << endl;
	}
	return 0;
}