wikioi 1035 火车停留 裸费用流

链接：http://wikioi.com/problem/1035/

怎么说呢，只能说这个建图很有意思。因为只有m条道，然后能互相接在一起的连通，对每个点进行拆点，很有意思的一道裸费用留题。

代码：

 #include <iostream>
 #include <queue>
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <vector>
 using namespace std;
 const int maxn = ;
 const int inf = ;
 struct node
 {
     int u,v,cap,flow,cost,next;
 }edges[];
 int head[maxn],cnt;
 void init(int n)
 {
     int i;
     for(i = ;i <= n;i++)
     head[i] = -;
     cnt = ;
 
     return ;
 }
 void addedge(int u,int v,int cap,int flow,int cost)
 {
     edges[cnt].u = u;
     edges[cnt].v = v;
     edges[cnt].cap = cap;
     edges[cnt].flow = flow;
     edges[cnt].cost = cost;
     edges[cnt].next = head[u];
     head[u] = cnt;
     cnt++;
     edges[cnt].u = v;
     edges[cnt].v = u;
     edges[cnt].cap = ;
     edges[cnt].flow = flow;
     edges[cnt].cost = -cost;
     edges[cnt].next = head[v];
     head[v] = cnt;
     cnt++;
 }
 int vis[maxn],a[maxn],pre[maxn],dis[maxn];
 int spfa(int s,int t,int n,int &flow,int &cost)
 {
     int i;
     queue<int> q;
     for(i = ;i <= n ;i++)
     dis[i] = -,vis[i] = ;
 
     dis[s] = ;
     pre[s] = ;
     vis[s] = ;
     a[s] = inf;
 
     int u,v;
     q.push(s);
 
     while(!q.empty())
     {
         u = q.front();
         q.pop();
         vis[u] = ;
 
         for(i = head[u];i != -;i = edges[i].next)
         {
             struct node & e = edges[i];
 
             v = e.v;
             if(e.cap > e.flow &&dis[v] < dis[u]+e.cost)
             {
                 dis[v] = dis[u]+e.cost;
                 a[v] = min(a[u],e.cap-e.flow);
                 pre[v] = i;
                 if(!vis[v])
                 {
                     vis[v] = ;
                     q.push(v);
                 }
             }
         }
     }
 
     if(dis[t] <= )
     return ;
     flow+= a[t];
     cost += dis[t]*a[t];
     u = t;
     while(u != s)
     {
         edges[pre[u]].flow += a[t];
         edges[pre[u]^].flow -= a[t];
         u = edges[pre[u]].u;
     }
     return ;
 }
 int MCMF(int s,int t,int n)
 {
     int flow = ,cost = ;
 
     while(spfa(s,t,n,flow,cost));
 
     return cost;
 }
 int r[],c[],s[];
 int main()
 {
     int n,m,u,v,i,j;
 
     scanf("%d %d",&n,&m);
     init(*m);
     for(i = ;i <= m;i++)
         cin>>r[i]>>c[i]>>s[i];
 
     addedge(,*m+,n,,);
     for(i = ;i <= m;i++)
     {
         addedge(*m+,i,inf,,);
         addedge(i,m+i,,,c[i]);
         addedge(m+i,*m+,inf,,);
         for(j = ;j <= m;j++)
         {
             if(r[i]+s[i] < r[j])
             addedge(m+i,j,inf,,);
         }
     }
     double ans;
     ans = 1.0*MCMF(,*m+,*m+);
 
     printf("%.2f\n",ans/);
     return ;
 }