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HDU3572_Task Schedule(网络流最大流)
阅读量:7127 次
发布时间:2019-06-28

本文共 3011 字,大约阅读时间需要 10 分钟。

解题报告

题意:

工厂有m台机器,须要做n个任务。对于一个任务i。你须要花费一个机器Pi天,并且,開始做这个任务的时间要>=Si,完毕这个任务的时间<=Ei。

对于一个任务,仅仅能由一个机器来完毕。一个机器同一时间仅仅能做一个任务。

可是,一个任务能够分成几段不连续的时间来完毕。问,是否能做完所有任务。

思路:

网络流在于建模,这题建模方式是:

把每一天和每一个任务看做点。由源点到每一任务。建容量为pi的边(表示任务须要多少天完毕)。

每一个任务每一天,若是能够在这天做任务,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 99999999using 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
 

转载地址:http://fahel.baihongyu.com/

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