hdu 3572 Task Schedule 网络流

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3572

Our geometry princess XMM has stoped her study in computational geometry to concentrate on her newly opened factory. Her factory has introduced M new machines in order to process the coming N tasks. For the i-th task, the factory has to start processing it at or after day Si, process it for Pi days, and finish the task before or at day Ei. A machine can only work on one task at a time, and each task can be processed by at most one machine at a time. However, a task can be interrupted and processed on different machines on different days.
Now she wonders whether he has a feasible schedule to finish
all the tasks in time. She turns to you for help.

 

题意：有M台机器和N个任务，每个任务必须在第si天到第ei天之间完成，完成一个任务需要pi天（1<=i<=n）。一台机器同时只能做一个任务，一个任务同一时间只能由一台机器处理。问能否完成。

 

解法：构造成为网络流，然后运用SAP算法求解。

把每个任务和每个时刻都作为节点，增设一个源点和一个汇点。对于任务i，连接一条源点指向 i 最大容量为pi 的边，然后连接 i 到[ si , ei ]区间里每个时刻，最大容量为1，最后连接每个时刻到汇点，最大容量为m（因为同一时刻最多只能有m台机器在操作）。此时，完成构图。

用网络流算法求解到最大流之后，判断是不是满流（每个任务都有天数的限制，不能少也不能多）。

SAP和Dinic算法：这道题里节点1000，边数50w，Dinic+邻接矩阵是不行的了，看到hdu上面有人800+msAC，也许Dinic+邻接表吧。我直接选用各种优化之后的SAP算法。

 

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<algorithm>
 #include<queue>
 #define inf 0x7fffffff
 using namespace std;
 const int maxn=+;
 const int M = +;

 struct Edge
 {
     int to,cap,next;
 }edge[M];
 int head[maxn],edgenum;
 int n,m,from,to,vnum;
 int level[maxn],gap[maxn];

 void add(int u,int v,int cap)
 {
     edge[edgenum].to=v;
     edge[edgenum].cap=cap;
     edge[edgenum].next=head[u];
     head[u]=edgenum++;

     edge[edgenum].to=u;
     edge[edgenum].cap=;
     edge[edgenum].next=head[v];
     head[v]=edgenum++;
 }

 void bfs(int to)
 {
     memset(level,-,sizeof(level));
     memset(gap,,sizeof(gap));
     level[to]=;
     gap[level[to] ]++;
     queue<int> Q;
     Q.push(to);
     while (!Q.empty())
     {
         int u=Q.front() ;Q.pop() ;
         for (int i=head[u] ;i!=- ;i=edge[i].next)
         {
             int v=edge[i].to;
             if (level[v] != -) continue;
             level[v]=level[u]+;
             gap[level[v] ]++;
             Q.push(v);
         }
     }
 }

 int pre[maxn];
 int cur[maxn];
 int SAP(int from,int to)
 {
     bfs(to);
     memset(pre,-,sizeof(pre));
     memcpy(cur,head,sizeof(head));
     int u=pre[from]=from,flow=,aug=inf;
     gap[]=vnum;
     while (level[from]<vnum)
     {
         bool flag=false;
         for (int &i=cur[u] ;i!=- ;i=edge[i].next)
         {
             int v=edge[i].to;
             if (edge[i].cap> && level[u]==level[v]+)
             {
                 flag=true;
                 pre[v]=u;
                 u=v;
                 aug=min(aug,edge[i].cap);
                 if (u==to)
                 {
                     flow += aug;
                     for (u=pre[u] ;v!=from ;v=u,u=pre[u])
                     {
                         edge[cur[u] ].cap -= aug;
                         edge[cur[u]^ ].cap += aug;
                     }
                     aug=inf;
                 }
                 break;
             }
         }
         if (flag) continue;
         int minlevel=vnum;
         for (int i=head[u] ;i!=- ;i=edge[i].next)
         {
             int v=edge[i].to;
             if (edge[i].cap> && level[v]<minlevel)
             {
                 minlevel=level[v];
                 cur[u]=i;
             }
         }
         if (--gap[level[u] ]==) break;
         level[u]=minlevel+;
         gap[level[u] ]++;
         u=pre[u];
     }
     return flow;
 }

 int main()
 {
     int t,ncase=;
     scanf("%d",&t);
     while (t--)
     {
         scanf("%d%d",&n,&m);
         int p,s,e;
         memset(head,-,sizeof(head));
         edgenum=;
         from=;
         int sn[],en[],maxe=;
         int sum=;
         for (int i= ;i<=n ;i++)
         {
             scanf("%d%d%d",&p,&s,&e);
             sum += p;
             sn[i]=s ;en[i]=e ;
             maxe=max(maxe,e);
             add(from,i,p);
             for (int j=s ;j<=e ;j++)
             {
                 add(i,n+j,);
             }
         }
         to=maxe+;
         vnum=to+;
         for (int j= ;j<=maxe ;j++)
             add(n+j,to,m);
         int Maxflow=SAP(from,to);
         printf("Case %d: ",ncase++);
         if (Maxflow==sum) printf("Yes\n");
         else printf("No\n");
         printf("\n");
     }
     return ;
 }