HDU4411 最小费用流

题目：http://acm.hdu.edu.cn/showproblem.php?pid=4411

1. floyd处理出最短路
2. 每个点拆为i、i+n，i到i+n连一条容量为1，费用为负无穷的边，代表这个城市必须访问
3. i+n到j（j>i)分别建边，容量为1，费用为i、j最短路
4. i+n到汇点建立容量为1费用为g[0][i]的边，代表会警察局
5. 0到i建立容量为1费用为g[0][i]的边
6. 0到汇点建立容量为k，费用为0的边，代表有的警察可能一直待在警察局
7. 源点到0建立容量为k费用为0的边

ps：这题会有重边，以后一定要注意，每道题都要判重！

 #include<bits/stdc++.h>
 using namespace std;
 const int inf = 0x3f3f3f3f;
 const int maxv = ;
 typedef pair<int,int> P;
 struct edge{
     int to,cap,cost,rev;
 };
 int V;
 vector<edge> g[maxv];
 int h[maxv];
 int dist[maxv];
 int prevv[maxv],preve[maxv];
 void addedge(int from,int to,int cap,int cost){
     g[from].push_back(edge{to,cap,cost,g[to].size()});
     g[to].push_back((edge{from,,-cost,g[from].size()-}));
 }
 int solve(int s,int t,int f){
     int res = ;
     memset(h,,sizeof(h)) ;
     while(f > ){
         priority_queue<P,vector<P>,greater<P> > que;
         memset(dist,inf,sizeof(dist));
         dist[s] = ;
         que.push(P(,s));
         while(!que.empty()){
             P p = que.top();
             que.pop();
             int v = p.second;
             if(dist[v] < p.first)
                 continue;
             for(int i = ;i<g[v].size();i++){
                 edge &e = g[v][i] ;
                 if(e.cap> && dist[e.to]>dist[v]+e.cost+h[v]-h[e.to]){
                     dist[e.to] = dist[v]+e.cost+h[v]-h[e.to];
                     prevv[e.to] = v;
                     preve[e.to] = i;
                     que.push(P(dist[e.to],e.to));
                 }
             }
         }
         if(dist[t] == inf)
             return -;
         for(int i = ;i<V;i++)
             h[i] += dist[i];
         int d = f;
         for(int v = t;v!=s;v=prevv[v])
             d = min(d,g[prevv[v]][preve[v]].cap);
         f -= d;
         res += d*h[t];
         for(int v = t;v!=s;v = prevv[v]){
             edge &e = g[prevv[v]][preve[v]];
             e.cap -= d;
             g[v][e.rev].cap += d;
         }
     }
     return res;
 }
 int mp[][];
 int main(){
     int n,m,k;
     while(scanf("%d%d%d",&n,&m,&k)){
         if(!n && !m && !k)
             break;
         V = *n+;
         int s = *n+,t = *n+;
         for(int i = ;i<*n+;i++)
             g[i].clear();
         memset(mp,inf,sizeof(mp));
         for(int i = ;i<=n;i++)
             mp[i][i] = ;
         int u,v,w;
         while(m--){
             scanf("%d%d%d",&u,&v,&w);
             mp[u][v] = mp[v][u] = min(mp[u][v],w);
         }
         for(int k = ;k<=n;k++)
             for(int i = ;i<=n;i++)
                 for(int j = ;j<=n;j++)
                     mp[i][j] = min(mp[i][j],mp[i][k]+mp[k][j]);
         for(int i = ;i<=n;i++){
             addedge(,i,,mp[][i]);
             addedge(i+n,t,,mp[][i]);
             addedge(i,i+n,,-);
             for(int j = i+;j<=n;j++){
                 if(mp[i][j] != inf)
                     addedge(i+n,j,k,mp[i][j]);
             }
         }
         addedge(s,,k,);
         addedge(,t,k,);
         cout << solve(s,t,k)+n* << endl;
     }
     return ;
 }