hdu 3572 Task Schedule
阿新 • • 發佈：2017-09-13
cout stdio.h res geometry stdin i++ rst math.h pro
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.
Task Schedule
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.
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.
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 題意：有n項任務和m臺機器，給出每個任務的執行時間pi，和開始時間si，以及截止時間ei。每項任務需要在si或si後開始執行，在ei或ei前執行完畢，每個任務可以分段，在任何一臺機器執行，每臺機器在一個時間只能執行一個任務。。 題解：拆點的最大流判斷是否滿流，建圖是重點。考慮每個任務可以在任何一臺機器執行，因此每個任務在規定的時間內就可以分到任何一臺機器（一條機器在一個時間點只執行一個任務），所以將時間進行拆點，添加一個源點S，從源點S向每個任務連邊，因為每個任務的執行時間是pi，所以從超級源點連向每個任務的流是pi，即建邊add_edge (S,i,pi)；然後考慮每個任務的執行，一個任務只能同時在一個時間點被工作，即不能同時既在時間點A上加工又在時間點B上加工，,所以每一個任務 i 向時間點[s,e]連一條容量為1的邊; add_edge (i,s--->t,1)；最後每一個時間點向T連一條容量為m個邊，說明一個時間點只能最多同時有m個機器在工作。最後判斷是否漫滿流，即判斷從S流出的
#include <stdio.h> #include <algorithm> #include <math.h> #include <string.h> #include <vector> #include <queue> #include <stack> #include <iostream> #define pi acos(-1.0) #define INF 0x3f3f3f3f using namespace std; #define ll long long const int maxm=251000+7; const int maxn=1010; struct edge { int t,w,f,next; }e[2*maxm]; int dis[maxn]; int head[maxn],cnt; void add_edge(int u,int v,int f) { e[cnt].t=v,e[cnt].f=f; e[cnt].next=head[u]; head[u]=cnt++; e[cnt].t=u,e[cnt].f=0; e[cnt].next=head[v]; head[v]=cnt++; } void init() { memset(head,-1,sizeof head); cnt=0; } int bfs(int t) { memset(dis,-1,sizeof dis); queue<int> q; dis[0]=1; q.push(0); while(!q.empty()) { int k=q.front();q.pop(); for(int i=head[k];i!=-1;i=e[i].next) { int v=e[i].t; if(e[i].f&&dis[v]==-1) { dis[v]=dis[k]+1; if(v==t) return 1; q.push(v); } } } return dis[t]!=-1; } int dfs(int s,int t,int f) { if(s==t||!f) return f; int r=0; for (int i=head[s]; i!=-1; i=e[i].next) { int v=e[i].t; if(dis[v]==dis[s]+1&&e[i].f) { int d=dfs(v,t,min(f,e[i].f)); if(d>0) { e[i].f-=d; e[i^1].f+=d; r+=d; f-=d; if(!f) break; } } } if(!r) dis[s]=INF; return r; } int main() { freopen("C:\\Users\\Administrator\\Desktop\\a.txt","r",stdin); //ios::sync_with_stdio(false); //freopen("C:\\Users\\Administrator\\Desktop\\b.txt","w",stdout); int T,n,m,Case=0; scanf("%d",&T); while(T--) { int st=INF,et=0,sump=0; scanf("%d%d",&n,&m); init(); for(int i=1;i<=n;i++) { int p,s,e; scanf("%d%d%d",&p,&s,&e); add_edge(0,i,p); sump+=p; st=min(st,s); et=max(et,e); for(int j=s;j<=e;j++) add_edge(i,n+j,1); } int t=n+et-st+2; //cout<<n+st<<" "<<n+et<<" "<<t<<endl; for(int i=st+n;i<=et+n;i++) add_edge(i,t,m); int ans=0,res; while(bfs(t)) { res=dfs(0,t,INF); ans+=res; } //printf("%d\n",ans); printf("Case %d: ",++Case); if(ans==sump) puts("Yes"); else puts("No"); printf("\n"); } return 0; }
hdu 3572 Task Schedule