HDU3572_Task Schedule(网络流最大流)
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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)。
每一个任务到每一天,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 <map> #include <queue> #include <vector> #include <cstdio> #include <cstring> #include <iostream> #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<int >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<e) maxx=e; for(j=s; j<=e; j++) add(i,j+n,1); } cnt=n+maxx+1; for(i=1; i<=maxx; i++) { add(n+i,cnt,m); } int ans=0,a; while(bfs()) while(a=dfs(0,inf)) ans+=a; printf("Case %d: ",k++); if(ans==sum) printf("Yes\n"); else printf("No\n"); printf("\n"); } return 0; }
Task Schedule
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3311 Accepted Submission(s): 1154
Now she wonders whether he has a feasible schedule to finish all the tasks in time. She turns to you for help.
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.
Print a blank line after each test case.
2 4 3 1 3 5 1 1 4 2 3 7 3 5 9 2 2 2 1 3 1 2 2
Case 1: Yes Case 2: Yes
field=problem&key=2010%20ACM-ICPC%20Multi-University%20Training%20Contest%A3%A813%A3%A9%A1%AA%A1%AAHost%20by%20UESTC&source=1&searchmode=source" style="color:rgb(26,92,200); text-decoration:none">2010 ACM-ICPC Multi-University Training Contest(13)——Host by UESTC
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