链接：http://acm.hdu.edu.cn/showproblem.php?pid=1853

Cyclic Tour

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)

Problem Description

There are N cities in our country, and M one-way roads connecting them. Now Little Tom wants to make several cyclic tours, which satisfy that, each cycle contain at least two cities, and each city belongs to one cycle exactly. Tom wants the total length of
all the tours minimum, but he is too lazy to calculate. Can you help him?

Input

There are several test cases in the input. You should process to the end of file (EOF).

The first line of each test case contains two integers N (N ≤ 100) and M, indicating the number of cities and the number of roads. The M lines followed, each of them contains three numbers A, B, and C, indicating that there is a road from city A to city B,
whose length is C. (1 ≤ A,B ≤ N, A ≠ B, 1 ≤ C ≤ 1000).

Output

Output one number for each test case, indicating the minimum length of all the tours. If there are no such tours, output -1.

Sample Input

6 9
1 2 5
2 3 5
3 1 10
3 4 12
4 1 8
4 6 11
5 4 7
5 6 9
6 5 4
6 5
1 2 1
2 3 1
3 4 1
4 5 1
5 6 1

Sample Output

42
-1

Hint

 In the first sample, there are two cycles, (1->2->3->1) and (6->5->4->6) whose length is 20 + 22 = 42. 

题意：

给你若干个点和带权有向边，要求把所有点连成环，可以多个环，但是每个环至少要有两个点。

做法：

所有的点成环，可以知道所有的点 入度和出度都为1。并且只要符合这个条件，所有点肯定是在一个环中的，也就是符合条件了。

所以可以建一个二分图，左边的点从s流入费用为0，流量为1，表示入度为1 ，右边一样。

然后根据边 建流量为1，费用为边权的边，这就是最大权值匹配的图了。

这样只要满流就符合条件了。

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
//最小费用最大流，求最大费用只需要取相反数，结果取相反数即可。
//点的总数为 N，点的编号 0~N-1
const int MAXN = 10000;
const int MAXM = 100000;
const int INF = 0x3f3f3f3f;
struct Edge
{
	int to,next,cap,flow,cost;
}edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;//节点总个数，节点编号从0~N-1
void init(int n)
{
	N = n;
	tol = 0;
	memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int cap,int cost)
{
	edge[tol].to = v;
	edge[tol].cap = cap;
	edge[tol].cost = cost;
	edge[tol].flow = 0;
	edge[tol].next = head[u];
	head[u] = tol++;
	edge[tol].to = u;
	edge[tol].cap = 0;
	edge[tol].cost = -cost;
	edge[tol].flow = 0;
	edge[tol].next = head[v];
	head[v] = tol++;
}
bool spfa(int s,int t)
{
	queue<int>q;
	for(int i = 0;i < N;i++)
	{
		dis[i] = INF;
		vis[i] = false;
		pre[i] = -1;
	}
	dis[s] = 0;
	vis[s] = true;
	q.push(s);
	while(!q.empty())
	{
		int u = q.front();
		q.pop();
		vis[u] = false;
		for(int i = head[u]; i != -1;i = edge[i].next)
		{
			int v = edge[i].to;
			if(edge[i].cap > edge[i].flow &&
				dis[v] > dis[u] + edge[i].cost )
			{
				dis[v] = dis[u] + edge[i].cost;
				pre[v] = i;
				if(!vis[v])
				{
					vis[v] = true;
					q.push(v);
				}
			}
		}
	}
	if(pre[t] == -1)return false;
	else return true;
}
//返回的是最大流， cost存的是最小费用
int minCostMaxflow(int s,int t,int &cost)
{
	int flow = 0;
	cost = 0;
	while(spfa(s,t))
	{
		int Min = INF;
		for(int i = pre[t];i != -1;i = pre[edge[i^1].to])
		{
			if(Min > edge[i].cap - edge[i].flow)
				Min = edge[i].cap - edge[i].flow;
		}
		for(int i = pre[t];i != -1;i = pre[edge[i^1].to])
		{
			edge[i].flow += Min;
			edge[i^1].flow -= Min;
			cost += edge[i].cost * Min;
		}
		flow += Min;
	}
	return flow;
}

int main()
{
	int n,m;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		init(2*n+2);
		int ss=0;
		int ee=2*n+1;
		for(int i=0;i<m;i++)
		{
			int u,v,w;
			scanf("%d%d%d",&u,&v,&w);

			addedge(u,v+n,1,w);

		}
		for(int i=1;i<=n;i++)
		{
			addedge(ss,i,1,0);
			addedge(i+n,ee,1,0);
		}

		int cost,liu;
		liu=minCostMaxflow(ss,ee,cost);
		if(liu!=n)
		{
			puts("-1");
		}
		else
		{
			printf("%d\n",cost);

		}

	}

	return 0;
}

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