POJ2135FarmTour（最小费用最大流模板题）

Farm Tour

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17273		Accepted: 6670

Description

When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M <= 10000) paths that connect the fields in various ways. Each path connects two different fields and has a nonzero length smaller than 35,000. 

To show off his farm in the best way, he walks a tour that starts at his house, potentially travels through some fields, and ends at the barn. Later, he returns (potentially through some fields) back to his house again. 

He wants his tour to be as short as possible, however he doesn't want to walk on any given path more than once. Calculate the shortest tour possible. FJ is sure that some tour exists for any given farm.

Input

* Line 1: Two space-separated integers: N and M. 

* Lines 2..M+1: Three space-separated integers that define a path: The starting field, the end field, and the path's length. 

Output

A single line containing the length of the shortest tour. 

Sample Input

4 5
1 2 1
2 3 1
3 4 1
1 3 2
2 4 2

Sample Output

6

Source

给出一个无向图，其中有n个点，m条边，要从1号区域走到n号区域，去的时候和回来的时候不能走一条重复的边。问这一去一回的最小花费。

思路：

首先是建边。因为是无向图，所以：

那么一去一回，其实可以考虑为两次去。

①每一次加入的边，都要双向建立，费用流沿用了最大流中的退回边，正向边建立的费用是w.退回边建立的费用是-w。因为要求的是每一条边只经过一次不重复，那么其流量限制就是1.

②建立S源点连入1号节点，没有花费，流量为2。表示要去两次N号节点。

②建立T汇点，从N号节点连入，同样没有花费，流量为2，表示要走到N号节点两次。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 1e3 + 7;
const int maxv = 200005;
const int INF = 0x3f3f3f3f;
int head[maxn], dis[maxn], path[maxn], pre[maxn], book[maxn] , n, m, s, t, k, sum;
struct node
{
    int v, w, f, next, cnt;
}edge[3000005];
void addEdge(int u, int v, int w, int f)
{
    edge[k].v = v;
    edge[k].w = w;
    edge[k].f = f;
    edge[k].cnt = k;
    edge[k].next = head[u];
    head[u] = k++;
    edge[k].v = u;
    edge[k].w = -w;
    edge[k].f = 0;
    edge[k].cnt = k;
    edge[k].next = head[v];
    head[v] = k++;
}
int spfa()
{
    queue q;
    q.push(s);
    memset(pre, -1, sizeof(pre));
    memset(path, -1, sizeof(path));
    for(int i = 1; i <= t; i++) dis[i] = INF;
    dis[s] = 0;
    memset(book, 0, sizeof(book));
    book[s] = 1;
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        book[u] = 0;
        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            int to = edge[i].v;
            int w = edge[i].w;
            int f = edge[i].f;
            if(f && dis[to] > dis[u] + w)
            {
                dis[to] = dis[u] + w;
                pre[to] = u;
                path[to] = edge[i].cnt;
                if(!book[to])
                {
                    q.push(to);
                    book[to] = 1;
                }
            }
        }
    }
    if(dis[t] != INF) return 1;
    else return 0;
}
void Min_costflow()
{
    int ans = 0;
    int maxflow = 0;
    while(spfa())
    {
        int minx = INF;
        for(int i = t; i != s; i = pre[i])
        {
            minx = min(minx, edge[path[i]].f);
        }
        maxflow += minx;
        ans += dis[t]*minx;
        for(int i = t; i != s; i = pre[i])
        {
            edge[path[i]].f -= minx;
            edge[path[i]^1].f += minx;
        }
    }
    printf("%d\n", ans);
}
int main()
{
    while(~scanf("%d%d", &n, &m))
    {
        s = 0, t = n+1, k = 0;
        memset(head, -1, sizeof(head));
        int x, y, z;
        for(int i = 1; i <= m; i++)
        {
            scanf("%d%d%d", &x, &y, &z);
            addEdge(x, y, z, 1);
            addEdge(y, x, z, 1);
        }
        addEdge(s, 1, 0, 2);
        addEdge(n, t, 0, 2);
        Min_costflow();
    }
    return 0;
}