BZOJ1070 修车-费用网络流

 1 #include<iostream>
 2 #include<vector>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<queue>
 6 using namespace std;
 7 const int maxn=3000;
 8 const int inf=1<<30;
 9 struct Edge{
10     int from,to,cap,flow,cost;
11 };
12 struct MCMF{
13     int n,m,s,t;
14     vector<Edge> edges;
15     vector<int> G[maxn];
16     int inq[maxn];
17     int d[maxn];
18     int p[maxn];
19     int a[maxn];
20     void init(int n){
21         this->n=n;
22         for(int i=1;i<=n;i++) G[i].clear();
23         edges.clear();
24     }
25     void addEdge(int from,int to,int cap,int cost){
26         edges.push_back((Edge){from,to,cap,0,cost});
27         edges.push_back((Edge){to,from,0,0,-cost});
28         m=edges.size();
29         G[from].push_back(m-2);
30         G[to].push_back(m-1);
31     }
32     bool BF(int s,int t,int& flow,int& cost){
33         for(int i=1;i<=n;i++) d[i]=inf;
34         memset(inq,0,sizeof(inq));
35         d[s]=0;inq[s]=1;p[s]=0;a[s]=inf;
36         
37         queue<int> Q;
38         Q.push(s);
39         while(!Q.empty()){
40             int u=Q.front();Q.pop();
41             inq[u]=0;
42             for(int i=0;i<G[u].size();i++){
43                 Edge& e=edges[G[u][i]];
44                 if(e.cap>e.flow&&d[e.to]>d[u]+e.cost){
45                     d[e.to]=d[u]+e.cost;
46                     p[e.to]=G[u][i];
47                     a[e.to]=min(a[u],e.cap-e.flow);
48                     if(!inq[e.to]){
49                         Q.push(e.to);
50                         inq[e.to]=1;
51                     }
52                 }
53             }
54         }
55         if(d[t]==inf) return false;
56         flow+=a[t];
57         cost+=d[t]*a[t];
58         int u=t;
59         while(u!=s){
60             edges[p[u]].flow+=a[t];
61             edges[p[u]^1].flow-=a[t];
62             u=edges[p[u]].from;
63         }
64         return true;
65     }
66     int MinCost(int s,int t){
67         int flow=0,cost=0;
68         while(BF(s,t,flow,cost));
69         return cost;
70     }
71 };
72 MCMF solver;
73 int t[61][10];
74 int main()
75 {
76     int n,m;
77     scanf("%d %d",&m,&n);
78     solver.init(2+n*(m+1));
79     for(int i=1;i<=n;++i)
80         for(int j=1;j<=m;++j)
81             scanf("%d",&t[i][j]);
82     for(int i=1;i<=n;++i)
83         solver.addEdge(1,1+i,1,0);
84     for(int i=1;i<=n*m;++i)
85         solver.addEdge(1+n+i,2+n+n*m,1,0);
86     for(int i=1;i<=n;i++){
87         for(int j=1;j<=m;j++){
88             int T=t[i][j];
89             for(int k=1;k<=n;k++)
90                 solver.addEdge(1+i,1+n+(j-1)*n+k,1,T*k);
91         }
92     }
93     printf("%.2lf",((double)solver.MinCost(1,2+n*(m+1)))/n);
94     return 0;
95 }