Cyclic Tour
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 3235 Accepted Submission(s): 1689
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
给你一个N个点M条边的带权有向图,现在要你求这样一个值:
该有向图中的所有顶点正好被1个或多个不相交的有向环覆盖(每个节点只能被一个有向环包含).
这个值就是 所有这些有向环的权值和. 要求该值越小越好.
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#define inf 1e9
#define maxn 205
using namespace std;
struct edge
{
int from,to,cap,flow,cost;
edge()
{
}
edge(int f,int t,int ca,int fl,int co):from(f),to(t),cap(ca),flow(fl),cost(co){}
};
struct mcnf
{
int n,m,s,t;
vector<edge>edges;
vector<int>g[maxn];
bool inq[maxn];
int d[maxn];
int p[maxn];
int a[maxn];
void init(int n,int s,int t)
{
this->n=n;
this->s=s;
this->t=t;
edges.clear();
for(int i=0;i<n;i++)
g[i].clear();
}
void addedge(int from,int to,int cap,int cost)
{
edges.push_back(edge(from,to,cap,0,cost));
edges.push_back(edge(to,from,0,0,-cost));
m=edges.size();
g[from].push_back(m-2);
g[to].push_back(m-1);
}
bool bellford(int &flow,int &cost)
{for(int i=0;i<n;i++)
d[i]=inf;
memset(inq,0,sizeof(inq));
d[s]=0;
a[s]=inf;
inq[s]=true;
p[s]=0;
queue<int>q;
q.push(s);
while(!q.empty())
{
int u=q.front();
q.pop();
inq[u]=false;
for(int i=0;i<g[u].size();i++)
{
edge&e=edges[g[u][i]];
if(e.cap>e.flow&&d[e.to]>d[u]+e.cost)
{
d[e.to]=d[u]+e.cost;
p[e.to]=g[u][i];
a[e.to]=min(a[u],e.cap-e.flow);
if(!inq[e.to])
{
q.push(e.to);
inq[e.to]=true;
}
}
}
}
if(d[t]==inf)
return false;
flow+=a[t];
cost+=a[t]*d[t];
int u=t;
while(u!=s)
{
edges[p[u]].flow+=a[t];
edges[p[u]^1].flow-=a[t];
u=edges[p[u]].from;
}
return true;
}
int mincost(int num)
{
int flow=0,cost=0;
while(bellford(flow,cost));
return flow==num?cost:-1;
}
}MM;
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
int src=0,dst=2*n+1;
MM.init(2*n+2,src,dst);
for(int i=1;i<=n;i++)
{MM.addedge(src,i,1,0);
MM.addedge(i+n,dst,1,0);
}
while(m--)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
MM.addedge(u,v+n,1,w);
}
printf("%d\n",MM.mincost(n));
}
return 0;
}