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?

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

[email protected]

Source

HDU 2007 Programming Contest - Final

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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*2];

int tope,sour=0,sink=1;

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=0;

        E[tope].c=-c;

        E[tope].next=V[y];

        V[y]=&E[tope];

E[tope].neg=&E[tope-1];

        E[tope-1].neg=&E[tope];

}

int q[MAXN*10],vis[MAXN],dis[MAXN],dfn=0;

int prev[MAXN];

Edge *path[MAXN];

int spfa()

{

        int ope=-1,clo=0,now;

        Edge *ne;

        memset(dis,INF,sizeof(dis));

        dfn++;

        q[0]=sour;

        vis[sour]=dfn;

        dis[sour]=0;

        while (ope<clo)

        {

                now=q[++ope];

                vis[now]=0;

                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=0;

        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=-1;

                memset(V,0,sizeof(V));

                for (i=0;i<n;i++)

                {

                        add_edge(sour,2+i,1,0);

                        add_edge(2+i+n,sink,1,0);

                }

                for(i=0;i<m;i++)

                {

                        scanf("%d%d%d",&x,&y,&z);x--;y--;

                        add_edge(2+x,2+y+n,1,z);

                }

                pair<int,int> p1;

                p1=max_cost_flow();

                if (p1.first!=n)

                {

                        printf("-1\n");

                }else

                {

                        printf("%d\n",p1.second);

                }

        }

}

Cyclic Tour HDUOJ 费用流