hdu 1853 Cyclic Tour【KM】

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

我的链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=17728#problem/C

KM算法模板：http://blog.csdn.net/cfreezhan/article/details/8246639

用KM求最小费用模板题：http://blog.csdn.net/cfreezhan/article/details/8256926

Cyclic Tour

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 857    Accepted Submission(s): 436

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. 
 

 

Author

RoBa@TJU

 

Source

HDU 2007 Programming Contest - Final

 

Recommend

lcy

题意：找出若干个环，覆盖所有的点，使得总花费最小。

思路：以城市的个数建图，初始化为最小值（大负数），注意重复路径，以长度的负值求KM，  

     从而求出最大匹配，返回负值即是最短路径。

//Accepted 300 KB 46 ms C++ 1463 B 2013-02-26 15:42:24

//Accepted	300 KB	46 ms	C++	1463 B	2013-02-26 15:42:24
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=110;
const int minn=-1<<30;
char map[maxn][maxn];
int w[maxn][maxn];
int lx[maxn],ly[maxn];//顶标  
bool s[maxn],t[maxn];//s[i]、t[i]为左/右第i个点是否已标记
int match[maxn];
int n;
int hungary(int u)//匈牙利,匹配（找增广路） 
{
	s[u]=true;//标记入匈牙利树 
	for(int v=1;v<=n;v++)
	{
		if(!t[v] && lx[u]+ly[v]==w[u][v])
		{
			t[v]=true;//标记入匈牙利树 易忘记 
			if(match[v]==-1 || hungary(match[v]))
			{
				match[v]=u;
				return true;
			}
		}
	}
	return false;
}
int KM()
{
	int sum=0;
	memset(match,-1,sizeof(match));//初始化,易写错 
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			lx[i]=-1<<30;//初始化 顶标 
	memset(ly,0,sizeof(ly));
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			lx[i]=max(lx[i],w[i][j]);//最大权 
	for(int i=1;i<=n;i++)//匹配每一个点  
	{
		while(true)
		{
			memset(s,false,sizeof(s));//每次匹配前，找增广路，初始化为0 
			memset(t,false,sizeof(t));
			if(hungary(i)) break;
			else{
				int a=1<<30;
				for(int j=1;j<=n;j++) if(s[j])//j匹配过（即在匈牙利树中） 
					for(int k=1;k<=n;k++) if(!t[k] && a>lx[j]+ly[k]-w[j][k])
						a=lx[j]+ly[k]-w[j][k];
				for(int j=1;j<=n;j++)//修改顶标 
				{
					if(s[j]) lx[j]-=a;
					if(t[j]) ly[j]+=a;
				}
			}
		}
	}
	for(int i=1;i<=n;i++) 
		sum+=w[match[i]][i];
	for(int i=1;i<=n;i++) 
		if(w[match[i]][i]==minn) return 1;//匹配不成功 
	return sum;
}
int main()
{
	int N,M;
	int u,v,length;
	while(scanf("%d%d",&N,&M)!=EOF)
	{
		n=N;
		for(int i=1;i<=N;i++)
			for(int j=1;j<=N;j++)
				w[i][j]=minn;
		for(int i=1;i<=M;i++)
		{
			scanf("%d%d%d",&u,&v,&length);
			if(-length>w[u][v]) w[u][v]=-length;
		}
		printf("%d\n",-KM());
	}
	return 0;
}