hdu 3572 Task Schedule (Dinic模板)

Problem Description

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.

 

Input

On the first line comes an integer T(T<=20), indicating the number of test cases.
You are given two integer N(N<=500) and M(M<=200) on the first line of each test case. Then on each of next N lines are three integers Pi, Si and Ei (1<=Pi, Si, Ei<=500), which have the meaning described in the description. It is guaranteed that in a feasible schedule every task that can be finished will be done before or at its end day.

 

Output

For each test case, print “Case x: ” first, where x is the case number. If there exists a feasible schedule to finish all the tasks, print “Yes”, otherwise print “No”.
Print a blank line after each test case.

 

Sample Input

2

4 3

1 3 5

1 1 4

2 3 7

3 5 9

2 2

2 1 3

1 2 2

 

Sample Output

Case 1: Yes

Case 2: Yes

 

给N个任务，M台机器。每个任务有最早才能开始做的时间S，deadline E，和持续工作的时间P。每个任务可以分段进行,但是在同一时刻,一台机器最多只能执行一个任务. 问存不存在可行的工作时间。

由于时间<=500且每个任务都能断断续续的执行,那么我们把每一天时间作为一个节点来用网络流解决该题.
建图: 源点s(编号0), 时间1-500天编号为1到500, N个任务编号为500+1 到500+N, 汇点t(编号501+N).
源点s到每个任务i有边(s, i, Pi)
每一天到汇点有边(j, t, M) (其实这里的每一天不一定真要从1到500,只需要取那些被每个任务覆盖的每一天即可)
如果任务i能在第j天进行,那么有边(i, j, 1) 注意由于一个任务在一天最多只有1台机器执行,所以该边容量为1,不能为INF或M哦.
最后看最大流是否 == 所有任务所需要的总天数.

 #include <bits/stdc++.h>
 
 using namespace std;
 const int maxn = ;
 const int inf = 0x3f3f3f3f;
 struct Edge
 {
     int from,to,cap,flow;
     Edge (int f,int t,int c,int fl)
     {
         from=f,to=t,cap=c,flow=fl;
     }
 };
 struct Dinic
 {
     int n,m,s,t;
     vector <Edge> edge;
     vector <int> G[maxn];//存图
     bool vis[maxn];//标记每点是否vis过
     int cur[maxn];//当前弧优化
     int dep[maxn];//标记深度
     void init(int n,int s,int t)//初始化
     {
         this->n=n;this->s=s;this->t=t;
         edge.clear();
         for (int i=;i<n;++i) G[i].clear();
     }
     void addedge (int from,int to,int cap)//加边,单向边
     {
         edge.push_back(Edge(from,to,cap,));
         edge.push_back(Edge(to,from,,));
         m=edge.size();
         G[from].push_back(m-);
         G[to].push_back(m-);
     }
     bool bfs ()
     {
         queue<int> q;
         while (!q.empty()) q.pop();
         memset(vis,false,sizeof vis);
         vis[s]=true;
         dep[s]=;
         q.push(s);
         while (!q.empty()){
             int u=q.front();
             //printf("%d\n",u);
             q.pop();
             for (int i=;i<G[u].size();++i){
                 Edge e=edge[G[u][i]];
                 int v=e.to;
                 if (!vis[v]&&e.cap>e.flow){
                     vis[v]=true;
                     dep[v]=dep[u]+;
                     q.push(v);
                 }
             }
         }
         return vis[t];
     }
     int dfs (int x,int mi)
     {
         if (x==t||mi==) return mi;
         int flow=,f;
         for (int &i=cur[x];i<G[x].size();++i){
             Edge &e=edge[G[x][i]];
             int y=e.to;
             if (dep[y]==dep[x]+&&(f=dfs(y,min(mi,e.cap-e.flow)))>){
                 e.flow+=f;
                 edge[G[x][i]^].flow-=f;
                 flow+=f;
                 mi-=f;
                 if (mi==) break;
             }
         }
         return flow;
     }
     int max_flow ()
     {
         int ans = ;
         while (bfs()){
             memset(cur,,sizeof cur);
             ans+=dfs(s,inf);
         }
         return ans;
     }
 }dinic;
 int full_flow;
 int main()
 {
     int casee = ;
     //freopen("de.txt","r",stdin);
     int T;scanf("%d",&T);
     while (T--){
         int n,m;
         full_flow = ;
         scanf("%d%d",&n,&m);
         int src = ,dst = ++n;
         dinic.init(++n,src,dst);
         bool vis[maxn];
         memset(vis,false,sizeof vis);
         for (int i=;i<=n;++i){
             int p,s,e;
             scanf("%d%d%d",&p,&s,&e);
             full_flow+=p;
             dinic.addedge(src,i+,p);
             for (int j=s;j<=e;++j){
                 vis[j]=true;
                 dinic.addedge(i+,j,);
             }
         }
         for (int i=;i<maxn;++i){
             if (vis[i])
                 dinic.addedge(i,dst,m);
         }
         printf("Case %d: ",++casee);
         if (dinic.max_flow()==full_flow){//dinic.max_flow()只能跑一遍
             printf("Yes\n\n");
         }
         else
             printf("No\n\n");
     }
     return ;
 }