Farm Tour

题意：一个人带领一位朋友游历他的牧场，他的房子在第一个点，仓库在第N个点。求解从第一个点出发到达第n个点后在返回，最短的路径，每个点只能走一次。

思路：费用流。建图：源点与第一个点建一条边花费2，流量0。第n个点与汇建立一条边，费用2，流量0，其他各点建图，费用1，流量为给出值。用MFMC模板求解。

#include<stdio.h>#include<string.h>#include<queue>#include<math.h>#include<iostream>using namespace std;#define inf 0xfffffff#define M 1005struct node{    int x, y;};int tem[M][M];int n, m, len;struct edge{    int v, w, flow, c, next;} edge[M * M * 10];int vst[M], dis[M], head[M], pre[M];//dis记录路径长度int tot, flow_sum;void add(int v, int w, int f, int c){    edge[tot].v = v;    edge[tot].w = w;    edge[tot].flow = f;    edge[tot].c = c;    edge[tot].next = head[v];    head[v] = tot++;    edge[tot].v = w;    edge[tot].w = v;    edge[tot].flow = 0;    edge[tot].c = -c;    edge[tot].next = head[w];    head[w] = tot++;}//MCMF开始bool spfa(int begin, int end)  //spfa求解最短路径，即增广链中最小费用{    int v, w;    queue < int >q;    for (int i = 0; i <= end + 2; i++)    {        pre[i] = -1;        vst[i] = 0;        dis[i] = inf;    }    vst[begin] = 1;    dis[begin] = 0;    q.push(begin);    while (!q.empty())    {        v = q.front();        q.pop();        vst[v] = false;        for (int i = head[v]; i != -1; i = edge[i].next)        {            if (edge[i].flow > 0)            {                w = edge[i].w;                if (dis[w] > dis[v] + edge[i].c)                {                    dis[w] = dis[v] + edge[i].c;                    pre[w] = i;                    if (!vst[w])                    {                        vst[w] = true;                        q.push(w);                    }                }            }        }    }    return dis[end] != inf;}int MCMF(int begin, int end){    int ans = 0, flow, i;    flow_sum = 0;    while (spfa(begin, end))    {        flow = inf;        for (i = pre[end]; i != -1; i = pre[edge[i].v])            if (edge[i].flow < flow)                flow = edge[i].flow;//找增广链上所有边的容量的最小值作为“可分配最大流”        for (i = pre[end]; i != -1; i = pre[edge[i].v])        {            edge[i].flow -= flow;    //正向弧加            edge[i^1].flow += flow;  //反向弧减        }        ans += dis[end];        flow_sum += flow; //总流量    }    // cout << flow_sum << endl;    return ans; // 返回最小费用}int main(){    int i, j;    int s,e,c;    while (scanf("%d%d", &n, &m) != EOF)    {        tot = 0;        memset(head, -1, sizeof (head));        for (i = 1; i <=m; i++)        {            cin>>s>>e>>c;            add(s,e,1,c);            add(e,s,1,c);        }        add(0, 1, 2, 0);        add(n, n+1, 2, 0);        cout << MCMF(0, n + 1) << endl;    }    return 0;}/*4 51 2 12 3 13 4 11 3 22 4 24 51 2 22 3 13 4 11 3 22 4 267*/