Cyclic Tour

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 2257    Accepted Submission(s): 1148

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.

题意:n个城市,m条路线,现在要对这些城市进行规划,每个城市都要被规划进一个环里面，而且只能出现一次,现在问能否做到?如果能够做到,输出最小路径和.

题解:最小费用最大流+拆点,如果城市在环里面,那么这个点的出度,入度都为1,将一个人拆成 i ,i+n，那么我们就限制了当前的人只能够用一次,然后对i ,j+n 建立关系,跑一遍最小费用最大流即可。如果最大流<n，那么则是无解的.

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int INF = 999999999;
const int N = 405;
const int M = 80005;
struct Edge{
    int u,v,cap,cost,next;
}edge[M];
int head[N],tot,low[N],pre[N];
int total ;
bool vis[N];
void addEdge(int u,int v,int cap,int cost,int &k){
    edge[k].u=u,edge[k].v=v,edge[k].cap = cap,edge[k].cost = cost,edge[k].next = head[u],head[u] = k++;
    edge[k].u=v,edge[k].v=u,edge[k].cap = 0,edge[k].cost = -cost,edge[k].next = head[v],head[v] = k++;
}
void init(){
    memset(head,-1,sizeof(head));
    tot = 0;
}
bool spfa(int s,int t,int n){
    memset(vis,false,sizeof(vis));
    for(int i=0;i<=n;i++){
        low[i] = (i==s)?0:INF;
        pre[i] = -1;
    }
    queue<int> q;
    q.push(s);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        vis[u] = false;
        for(int k=head[u];k!=-1;k=edge[k].next){
            int v = edge[k].v;
            if(edge[k].cap>0&&low[v]>low[u]+edge[k].cost){
                low[v] = low[u] + edge[k].cost;
                pre[v] = k; ///v为终点对应的边
                if(!vis[v]){
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
    if(pre[t]==-1) return false;
    return true;
}
int MCMF(int s,int t,int n){
    int mincost = 0,minflow,flow=0;
     while(spfa(s,t,n))
    {
        minflow=INF+1;
        for(int i=pre[t];i!=-1;i=pre[edge[i].u])
            minflow=min(minflow,edge[i].cap);
        flow+=minflow;
        for(int i=pre[t];i!=-1;i=pre[edge[i].u])
        {
            edge[i].cap-=minflow;
            edge[i^1].cap+=minflow;
        }
        mincost+=low[t]*minflow;
    }
    total=flow;
    return mincost;
}
int n,m;
int main(){
     while(scanf("%d%d",&n,&m)!=EOF){
        init();
        int src = 0,des = 2*n+1;
        for(int i=1;i<=n;i++){
            addEdge(src,i,1,0,tot);
            addEdge(i+n,des,1,0,tot);
        }
        for(int i=1;i<=m;i++){
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            addEdge(u,v+n,1,w,tot);
        }
        int mincost = MCMF(src,des,2*n+2);
        if(total!=n) printf("-1\n");
        else printf("%d\n",mincost);
     }
}