BZOJ 1070: [SCOI2007]修车（费用流）

http://www.lydsy.com/JudgeOnline/problem.php?id=1070

题意：

思路：

神奇的构图。

因为排在后面的人需要等待前面的车修好，这里将每个技术人员拆成n个点，第k个点表示该技术人员倒数第k的顺序来修理该车，此时它的时间对于答案的贡献就是kw。

最后跑一遍最小费用流。

 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<cstdio>
 #include<vector>
 #include<stack>
 #include<queue>
 #include<cmath>
 #include<map>
 #include<set>
 using namespace std;
 typedef long long ll;
 typedef pair<int,int> pll;
 const int INF = 0x3f3f3f3f;
 const int maxn =  + ;

 int n,m;

 struct Edge
 {
     int from, to, cap, flow, cost;
     Edge(int u, int v, int c, int f, int w) :from(u), to(v), cap(c), flow(f), cost(w) {}
 };

 struct MCMF
 {
     int n, m;
     vector<Edge> edges;
     vector<int> G[maxn];
     int inq[maxn];
     int d[maxn];
     int p[maxn];
     int a[maxn];

     void init(int n)
     {
         this->n = n;
         for (int i = ; i<n; i++) G[i].clear();
         edges.clear();
     }

     void AddEdge(int from, int to, int cap, int cost)
     {
         edges.push_back(Edge(from, to, cap, , cost));
         edges.push_back(Edge(to, from, , , -cost));
         m = edges.size();
         G[from].push_back(m - );
         G[to].push_back(m - );
     }

     bool BellmanFord(int s, int t, int &flow, int & cost)
     {
         for (int i = ; i<n; i++) d[i] = INF;
         memset(inq, , sizeof(inq));
         d[s] = ; inq[s] = ; p[s] = ; a[s] = INF;

         queue<int> Q;
         Q.push(s);
         while (!Q.empty()){
             int u = Q.front(); Q.pop();
             inq[u] = ;
             for (int i = ; i<G[u].size(); i++){
                 Edge& e = edges[G[u][i]];
                 if (e.cap>e.flow && d[e.to]>d[u] + e.cost){
                     d[e.to] = d[u] + e.cost;
                     p[e.to] = G[u][i];
                     a[e.to] = min(a[u], e.cap - e.flow);
                     if (!inq[e.to]) { Q.push(e.to); inq[e.to] = ; }
                 }
             }
         }

         if (d[t] == INF) return false;
         flow += a[t];
         cost += d[t] * a[t];
         for (int u = t; u != s; u = edges[p[u]].from)
         {
             edges[p[u]].flow += a[t];
             edges[p[u] ^ ].flow -= a[t];
         }
         return true;
     }

     int MincostMaxdflow(int s, int t){
         int flow = , cost = ;
         while (BellmanFord(s, t, flow, cost));
         return cost;
     }
 }t;

 int main()
 {
     //freopen("in.txt","r",stdin);
     while(~scanf("%d%d",&m,&n))
     {
         int src=,dst=n*m+n+;
         t.init(dst+);
         for(int i=;i<=n;i++)
         {
             t.AddEdge(src,i,,);
             for(int j=;j<=m;j++)
             {
                 int x; scanf("%d",&x);
                 for(int k=;k<=n;k++)
                     t.AddEdge(i,j*n+k,,k*x);
             }
         }
         for(int j=n+;j<=m*n+n;j++)
             t.AddEdge(j,dst,,);
         int ans = t.MincostMaxdflow(src,dst);
         printf("%.2f\n",(double)ans/n);
     }
     return ;
 }