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HDU3572_Task Schedule(网络流最大流)
2015-07-20 18:03:12 】 浏览:3763
Tags:HDU3572_Task Schedule 网络 最大

解题报告

题意:

工厂有m台机器,需要做n个任务。对于一个任务i,你需要花费一个机器Pi天,而且,开始做这个任务的时间要>=Si,完成这个任务的时间<=Ei。对于一个任务,只能由一个机器来完成,一个机器同一时间只能做一个任务。但是,一个任务可以分成几段不连续的时间来完成。问,能否做完全部任务。

思路:

网络流在于建模,这题建模方式是:

把每一天和每个任务看做点。由源点到每一任务,建容量为pi的边(表示任务需要多少天完成)。每个任务到每一天,若是可以在这天做任务,建一条容量为1的边,最后,把每天到汇点再建一条边容量m(表示每台机器最多工作m个任务)。

#include 
#include 
   
     #include 
    
      #include 
     
       #include 
      
        #include 
       
         #define inf 99999999 using namespace std; int n,m,l[2010],head[2010],cnt,M; struct node { int v,w,next; } edge[555000]; void add(int u,int v,int w) { edge[M].v=v; edge[M].w=w; edge[M].next=head[u]; head[u]=M++; edge[M].v=u; edge[M].w=0; edge[M].next=head[v]; head[v]=M++; } int bfs() { memset(l,-1,sizeof(l)); l[0]=0; int i,u,v; queue
        
         Q; Q.push(0); while(!Q.empty()) { u=Q.front(); Q.pop(); for(i=head[u]; i!=-1; i=edge[i].next) { v=edge[i].v; if(l[v]==-1&&edge[i].w>0) { l[v]=l[u]+1; Q.push(v); } } } if(l[cnt]>0)return 1; return 0; } int dfs(int u,int f) { int a,i; if(u==cnt)return f; for(i=head[u]; i!=-1; i=edge[i].next) { int v=edge[i].v; if(l[v]==l[u]+1&&edge[i].w>0&&(a=dfs(v,min(f,edge[i].w)))) { edge[i].w-=a; edge[i^1].w+=a; return a; } } l[u]=-1;//没加优化会T return 0; } int main() { int t,i,j,s,p,e,k=1; scanf("%d",&t); while(t--) { M=0; memset(head,-1,sizeof(head)); scanf("%d%d",&n,&m); int sum=0,maxx=0; for(i=1; i<=n; i++) { scanf("%d%d%d",&p,&s,&e); add(0,i,p); sum+=p; if(maxx
         
          

Task Schedule

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3311 Accepted Submission(s): 1154


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

Author allenlowesy
Source 2010 ACM-ICPC Multi-University Training Contest(13)――Host by UESTC

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