hdu-3572 Task Schedule---Maximum flow estimation +dinic algorithm

Topic Links:

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

Main topic:

Give n a task, M machine. Each task has the earliest time to start doing s,deadline E, and the duration of the work is P. Each task can be segmented, but at the same time, a single machine can perform at most one task. There is no viable working time to ask for a deposit.

Problem Solving Ideas:

Since time <=500 and each task can be executed intermittently, we use the network stream as a node to solve the problem.

Map: Source point S (number 0), time 1-500 days numbered 1 to ten, N tasks numbered 500+1 to 500+n, meeting point T (numbered 501+n).

Source point s to each task I have side (s, I, Pi)

Every day to the meeting point there is an edge (J, T, M) (in fact, every day here is not necessarily from 1 to 500, just take those that are covered by each task each day)

If task I can be performed on day J, then there is an edge (I, J, 1) note that because a task is executed at most 1 machines in a day, the edge capacity is 1 and cannot be inf or M oh.

Finally, see if the maximum flow is = = Total number of days required for all tasks.

The maximum flow of the augmented path will time out, and the augmented path should be optimized with the dinic algorithm.

(The topic of graph theory is difficult to figure out how to build, not the use of algorithms)

1#include <bits/stdc++.h>2 using namespacestd;3typedefLong Longll;4 Const intMAXN = ++Ten;5 Const intINF =1e9;6 structEdge7 {8     intu, V, C, F;9 Edge () {}TenEdgeintUintVintCintf): U (U), V (v), C (c), F (f) {} One }; AVector<edge>e; -vector<int>G[MAXN]; - intLEVEL[MAXN];//BFS tiering the intITER[MAXN];//Current ARC Optimization - intN, M; - voidInit () - { +      for(inti =0; i < MAXN; i++) g[i].clear (); - e.clear (); + } A voidAddedge (intUintVintc) at { -E.push_back (Edge (U, V, C,0)); -E.push_back (Edge (V, U,0,0)); -     intm =e.size (); -G[u].push_back (M-2); -G[v].push_back (M-1); in } - voidBFS (ints) to { +Memset (Level,-1,sizeof(level)); -queue<int>Q; theLevel[s] =0; * Q.push (s); $      while(!q.empty ())Panax Notoginseng     { -         intU =Q.front (); the Q.pop (); +          for(inti =0; I < g[u].size (); i++) A         { theedge& now =E[g[u][i]]; +             intv =now.v; -             if(Now.c > Now.f && level[v] <0) $             { $LEVEL[V] = Level[u] +1; - Q.push (v); -             } the         } -     }Wuyi } the intDfsintUintTintF//U is the current point, T is the end, and F is the current flow - { Wu     if(U = = t)returnF; -      for(int& i = Iter[u]; I < g[u].size (); i++)//Current ARC Optimization About     { $edge& now =E[g[u][i]]; -         intv =now.v; -         if(Now.c > Now.f && level[u] <Level[v]) -         { A             intD = Dfs (V, t, Min (F, now.c-now.f)); +             if(D >0) the             { -NOW.F + =D; $E[g[u][i] ^1].F-=D; the                 returnD; the             } the         } the     } -     return 0; in } the intMaxflow (intSintt) the { About     intFlow =0; the      for(;;) the     { the BFS (s); +         if(Level[t] <0)returnflow; -memset (ITER,0,sizeof(ITER)); the         intF;Bayi          while(f = DFS (S, T, INF)) >0) Flow + =F; the     } the     returnflow; - } - intMain () the { the     intT, cases =0; thescanf"%d", &T); the      while(t--) -     { thescanf"%d%d", &n, &m); the         intMaxday =0; the         intSumday =0;94 init (); the         intp, S, E; the          for(inti =1; I <= N; i++) the         {98scanf"%d%d%d", &p, &s, &e); AboutAddedge (0, -+I, p); -              for(intj = S; J <= E; J + +)101Addedge (i + -J1);102Maxday =Max (Maxday, e);103Sumday + =p;104         } the          for(inti =1; I <= maxday; i++)106Addedge (I,501+N, m);107         if(Maxflow (0,501+ N) = =sumday)108printf"Case %d:yes\n\n", ++cases);109         Elseprintf"Case %d:no\n\n", ++cases); the     }111     return 0; the}

hdu-3572 Task Schedule---Maximum flow estimation +dinic algorithm