HDU3488 Tour [有向环覆盖 费用流]

Tour

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 3159    Accepted Submission(s): 1525

Problem Description

In the kingdom of Henryy, there are N (2 <= N <= 200) cities, with M (M <= 30000) one-way roads connecting them. You are lucky enough to have a chance to have a tour in the kingdom. The route should be designed as: The route should contain one or more loops. (A loop is a route like: A->B->……->P->A.)
Every city should be just in one route.
A loop should have at least two cities. In one route, each city should be visited just once. (The only exception is that the first and the last city should be the same and this city is visited twice.)
The total distance the N roads you have chosen should be minimized.

 

Input

An integer T in the first line indicates the number of the test cases.
In each test case, the first line contains two integers N and M, indicating the number of the cities and the one-way roads. Then M lines followed, each line has three integers U, V and W (0 < W <= 10000), indicating that there is a road from U to V, with the distance of W.
It is guaranteed that at least one valid arrangement of the tour is existed.
A blank line is followed after each test case.

 

Output

For each test case, output a line with exactly one integer, which is the minimum total distance.

 

Sample Input

1 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

 

Sample Output

42

 

Source

2010 ACM-ICPC Multi-University Training Contest（6）——Host by BIT

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和DAG的最小路径覆盖很像 http://www.cnblogs.com/candy99/p/6115989.html

也是拆入点和出点建二分图，然后求最小权最大匹配，因为保证有解所以一定是一个完美匹配，这个费用就是答案了

 

此题时间太烦了，必须要先去重边..............................

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
const int N=505,M=1e5,INF=1e9;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}
int n,m,u,v,w,s,t;
struct edge{
    int v,ne,c,f,w;
}e[M<<1];
int cnt,h[N];
inline void ins(int u,int v,int c,int w){
    cnt++;
    e[cnt].v=v;e[cnt].c=c;e[cnt].f=0;e[cnt].w=w;
    e[cnt].ne=h[u];h[u]=cnt;
    cnt++;
    e[cnt].v=u;e[cnt].c=0;e[cnt].f=0;e[cnt].w=-w;
    e[cnt].ne=h[v];h[v]=cnt;
}
int d[N],pre[N],pos[N],q[N],head=1,tail=1,inq[N];
inline void lop(int &x){if(x==N) x=1;else if(x==0) x=N-1;}
bool spfa(){
    memset(d,127,sizeof(d));
    d[s]=0;pre[t]=-1;
    head=tail=1;
    memset(inq,0,sizeof(inq));
    q[tail++]=s;inq[s]=1;
    while(head!=tail){
        int u=q[head++];lop(head);inq[u]=0;
        for(int i=h[u];i;i=e[i].ne){
            int v=e[i].v,w=e[i].w;
            if(d[v]>d[u]+w&&e[i].c>e[i].f){
                d[v]=d[u]+w;
                pre[v]=u;
                pos[v]=i;
                if(!inq[v]){
                    if(d[v]<d[q[head]]) head--,lop(head),q[head]=v;
                    else q[tail++]=v,lop(tail);
                    inq[v]=1;
                }
            }
        }
    }
    return pre[t]==-1?0:1;
}
int mcmf(){
    int flow=0,cost=0;
    while(spfa()){
        int f=INF;
        for(int i=t;i!=s;i=pre[i]) f=min(f,e[pos[i]].c-e[pos[i]].f);
        flow+=f;
        cost+=f*d[t];
        for(int i=t;i!=s;i=pre[i]){
            e[pos[i]].f+=f;
            e[((pos[i]-1)^1)+1].f-=f;
        }
    }
    return cost;
}

int g[N][N];
int main(){
    //freopen("in.txt","r",stdin);
    int T=read();
while(T--){
    cnt=0;
    memset(h,0,sizeof(h));
    n=read();m=read();s=0;t=n+n+1;
    for(int i=1;i<=n;i++){
        ins(s,i,1,0),ins(n+i,t,1,0);
        for(int j=1;j<=n;j++) g[i][j]=INF;
    }
    
    for(int i=1;i<=m;i++) u=read(),v=read(),w=read(),g[u][v]=min(g[u][v],w);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++) 
            if(g[i][j]<INF) ins(i,j+n,1,g[i][j]);
    printf("%d\n",mcmf());
}
}

 

 

 

转载于:https://www.cnblogs.com/candy99/p/6347825.html