8.15 M - Cyclic Tour

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本文介绍了一种利用最小费用最大流算法解决特定旅行问题的方法。该问题要求找到连接多个城市的最优环路组合，确保每个城市恰好被一个环路覆盖，并使所有环路的总长度最短。

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M - Cyclic Tour

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

Input

There are several test cases in the input. You shouldprocess 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 aroad 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 theminimum 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

-1

Hint

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

题意描述：n个城市里有m条单向路径，每条路径上有一个权值，每个城市都属于且仅属于某一个环，Tom计划环游这n个城市，并且每个城市都只能经过一次。问最后环游了n个城市后最小的权值和是多少。

思路：这道题是一道最小费用最大流问题，先将每条边权值负数存储，用km算法求最小权，跟最大权一样，只是建图的时候权值取负，求出的最大权绝对值是最小的。取反回来后，就是所求的解。同时，还要判断是否将所有路径全部加入，如果某个点没有构成回路，则说明无法遍历全部城市，输出-1。

#include<stdio.h>
#include<string.h>
int a[120][120],x1[120],y1[120],x2[120],y2[120],match[120],c[120];
int m,n;
int max(int x,int y)
{
	if(x<y)
		x=y;
	return x;
}
int min(int x,int y)
{
	if(x>y)
		x=y;
	return x;
}


int dfs(int w)
{
	int i,t;
	x2[w]=1;
	for(i=1;i<=n;i++)
	{
		if(y2[i])
			continue;
		t=x1[w]+y1[i]-a[w][i];
		if(t==0)
		{
			y2[i]=1;
			if(match[i]==-1||dfs(match[i]))
			{
				match[i]=w;
				return 1;
			}
		}
		else
		{
			c[i]=min(c[i],t);
		}
	}
	return 0;
}

void km()
{
	int i,j,t;
	memset(match,-1,sizeof(match));
	memset(y1,0,sizeof(y1));
	for(i=1;i<=n;i++)
	{
		x1[i]=a[i][1];
		for(j=2;j<=n;j++)
		{
			x1[i]=max(x1[i],a[i][j]);
		}
	}
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=n;j++)
			c[j]=1e9;
		while(1)
		{
			memset(x2,0,sizeof(x2));
			memset(y2,0,sizeof(y2));
			if(dfs(i))
				break;
			t=1e9;
			for(j=1;j<=n;j++)
			{
				if(!y2[j]&&t>c[j])
					t=c[j];
			}
			for(j=1;j<=n;j++)
			{
				if(x2[j])
					x1[j]-=t;
			}
			for(j=1;j<=n;j++)
			{
				if(y2[j])
					y1[j]+=t;
				else
					c[j]-=t;
			}
		}
	}
	int sum=0,f=0;
	for(i=1;i<=n;i++)
	{
		if(match[i]==-1||a[match[i]][i]==-1e9)//注意判断是否将全部的点加入
		{
			f=1;
			break;
		}
		sum+=a[match[i]][i];
	}		
	if(f==1)
	{
		printf("-1\n");
	}
	else
		printf("%d\n",-sum);
}
int main()
{
	int i,j,k;
	int t1,t2,t3;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
				a[i][j]=-1e9;
		}
		for(i=1;i<=m;i++)
		{
			scanf("%d%d%d",&t1,&t2,&t3);
			if(t3<-a[t1][t2])
			{
				a[t1][t2]=-t3;
			}
		}
	/*	for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
			{
				printf("%d  ",a[i][j]);
			}
			printf("\n");
		}
	*/	km();
	}
	return 0;
}