poj 2135最小费用最大流模板

最小费用最大流模板，这是摘抄别人的模板、、、罪恶啊，等省赛后好好搞网络流自己敲。

题意给出一个无向图，求1到n点最短环路，要求每个点只能经过一次；

思路：即找出图中无交集的两条最短路，建图，源点与1，汇点与n相连，同时附容量为2，其他边则附容量为1。

代码:

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
#define typef int
#define typec int
#define maxn 1005
#define maxm 10005
#define N maxn + 2
#define E maxm * 4 + 4
const typef inff = 0x3f3f3f3f;
const typec infc = 0x3f3f3f3f;

struct network
{
int nv, ne, pnt[E], nxt[E];
int vis
, que
, head
, pv
, pe
;
typef flow, cap[E];
typec cost, dis[E], d
;
void addedge(int u, int v, typef c, typec w)
{
pnt[ne] = v;
cap[ne] = c;
dis[ne] = +w;
nxt[ne] = head[u];
head[u] = (ne++);
pnt[ne] = u;
cap[ne] = 0;
dis[ne] = -w;
nxt[ne] = head[v];
head[v] = (ne++);
}
int mincost(int src, int sink)
{
int i, k, f, r;
typef mxf;
for (flow = 0, cost = 0;;)
{
memset(pv, -1, sizeof(pv));
memset(vis, 0, sizeof(vis));
for (i = 0; i < nv; ++i)
d[i] = infc;
d[src] = 0;
pv[src] = src;
vis[src] = 1;
for (f = 0, r = 1, que[0] = src; r != f;)
{
i = que[f++];
vis[i] = 0;
if (N == f)
f = 0;
for (k = head[i]; k != -1; k = nxt[k])
if (cap[k] && dis[k] + d[i] < d[pnt[k]])
{
d[pnt[k]] = dis[k] + d[i];
if (0 == vis[pnt[k]])
{
vis[pnt[k]] = 1;
que[r++] = pnt[k];
if (N == r)
r = 0;
}
pv[pnt[k]] = i;
pe[pnt[k]] = k;
}
}
if (-1 == pv[sink])
break;
for (k = sink, mxf = inff; k != src; k = pv[k])
if (cap[pe[k]] < mxf)
mxf = cap[pe[k]];
flow += mxf;
cost += d[sink] * mxf;
for (k = sink; k != src; k = pv[k])
{
cap[pe[k]] -= mxf;
cap[pe[k] ^ 1] += mxf;
}
}
return cost;
}
void build(int v, int e)
{
nv = v;
ne = 0;
memset(head, -1, sizeof(head));
int x, y;
typec w;
for (int i = 0; i < e; ++i)
{
scanf("%d%d%d", &x, &y, &w);
addedge(x, y, 1, w);// add arc (u->v, f, w)
addedge(y, x, 1, w);
}
addedge(0, 1, 2, 0);
addedge(v - 2, v - 1, 2, 0);
}
} g;

int n, m;

int main()
{
//freopen("t.txt", "r", stdin);
scanf("%d%d", &n, &m);
g.build(n + 2, m);
printf("%d\n", g.mincost(0, n + 1));
return 0;
}