最小费用最大流

#include<iostream>
using namespace std;
struct edge
{
	int u, v, next, f, w;
}e[40000001];
int n, m;
int mincost;
int p[200000];
int head[200000];
int num;
int s, t;
int max_flow;
void add(int s, int t, int w, int c)
{
	e[num].f = c;
	e[num].next = head[s];
	head[s] = num;
	e[num].u = s;
	e[num].v = t;
	e[num++].w = w;
	return ;
}
void addedge(int s, int t, int w, int c)
{
	add(s, t, w, c);
	add(t, s, -w, 0);
	return ;
}
int Q[200000]/*开大点*/,d[200000];
bool spfa()
{
	int i, j;
	int front = 0, tail = 0;
	bool ok[200000] = {0};//注意不
	for(i = 0; i <= t; i++)//t一般需要是最大的值，
	{
		d[i] = INT_MAX;
	}
	d[s] = 0;
	ok[s] = 1;
	p[s] = -1;
	Q[++tail] = s;
	while(tail! = front)
	{
		i = Q[++front];
		ok[i] = 0;
		for(j = head[i]; j != -1; j = e[j].next)
		{
			if(e[j].f > 0 && d[i] + e[j].w < d[e[j].v])
			{
				d[e[j].v] = d[i] + e[j].w;
				p[e[j].v] = j;
				if(!ok[e[j].v])
				{
					Q[++tail] = e[j].v;
					ok[e[j].v] = 1;
				}
			}
		}
	}
	if(d[t] < INT_MAX)
		return 1;
	else return 0;
}
int solve()
{
	int i;
	int minflow = INT_MAX;
	for(i = p[t]; i != -1; i = p[e[i].u])
	{
		if(minflow > e[i].f)
			minflow = e[i].f;
	}
	for (i = p[t]; i != -1; i = p[e[i].u])
	{
		e[i].f -= minflow;
		e[i^1].f += minflow;
		mincost += e[i].w*minflow;
	}
	max_flow += minflow;
	return 0;
}
/*
int main()
{

	memset(head, -1, sizeof(head));
	num = 0;
	s = 0;
	t = n + 1;
	max_flow = 0;
	//如果有a->b的边，容量为w,单位费用为c则：
	addedge(a, b, c, w);
	mincost = 0;
	while(spfa())
	{
		solve();
	}
	printf("%d\n",mincost);
}
return 0;
}
*/
void init()
{

	memset(head, -1, sizeof(head));
	num = 0;
//	s=0; 
//	t=n+1;
	max_flow = 0;
	//如果有a->b的边，容量为w,单位费用为c则：
//	addedge(a, b, c, w);
	mincost=0;
}