Cyclic Tour
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 1399 Accepted Submission(s): 712
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?
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).
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
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
-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
Recommend
费用流，拆点，source连接每个入点，费用0，容量1，每个出点连sink，费用0，容量1，入点出点见连费用边权，容量1，裸的费用流。
只要注意一下spfa的队列数组开大一点。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define MAXN 210
#define INF 0x3f3f3f3f
//AC
int n,m;
struct Edge
{
int np,val,c;
Edge *next,*neg;
}E[MAXN*MAXN],*V[MAXN*];
int tope,sour=,sink=;
void add_edge(int x,int y,int z,int c)
{
//cout<<"Add"<<x<<" "<<y<<" "<<z<<" "<<c<<endl;
E[++tope].np=y;
E[tope].val=z;
E[tope].c=c;
E[tope].next=V[x];
V[x]=&E[tope];
E[++tope].np=x;
E[tope].val=;
E[tope].c=-c;
E[tope].next=V[y];
V[y]=&E[tope]; E[tope].neg=&E[tope-];
E[tope-].neg=&E[tope];
}
int q[MAXN*],vis[MAXN],dis[MAXN],dfn=;
int prev[MAXN];
Edge *path[MAXN];
int spfa()
{
int ope=-,clo=,now;
Edge *ne;
memset(dis,INF,sizeof(dis));
dfn++;
q[]=sour;
vis[sour]=dfn;
dis[sour]=;
while (ope<clo)
{
now=q[++ope];
vis[now]=;
for (ne=V[now];ne;ne=ne->next)
{
if (ne->val&&dis[ne->np]>dis[now]+ne->c)
{
dis[ne->np]=dis[now]+ne->c;
prev[ne->np]=now;
path[ne->np]=ne;
if (vis[ne->np]!=dfn)
{
vis[ne->np]=dfn;
q[++clo]=ne->np;
}
}
}
}
return dis[sink];
}
pair<int,int> max_cost_flow()
{
int ds,fl,now,x;
pair<int,int> ret;
ret.first=ret.second=;
while (ds=spfa(),ds!=INF)
{
x=sink;
fl=INF;
while (x!=sour)
{
fl=min(fl,path[x]->val);
x=prev[x];
}
x=sink;
while (x!=sour)
{
path[x]->val-=fl;
path[x]->neg->val+=fl;
x=prev[x];
}
ret.first+=fl;
ret.second+=ds*fl;
}
return ret;
}
int main()
{
freopen("input.txt","r",stdin);
int i,j,k,x,y,z;
while (~scanf("%d%d",&n,&m))
{
tope=-;
memset(V,,sizeof(V));
for (i=;i<n;i++)
{
add_edge(sour,+i,,);
add_edge(+i+n,sink,,);
}
for(i=;i<m;i++)
{
scanf("%d%d%d",&x,&y,&z);x--;y--;
add_edge(+x,+y+n,,z);
}
pair<int,int> p1;
p1=max_cost_flow();
if (p1.first!=n)
{
printf("-1\n");
}else
{
printf("%d\n",p1.second);
}
}
}