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[N], que[N], head[N], pv[N], pe[N];
    typef flow, cap[E];
    typec cost, dis[E], d[N];
    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;
}