Hdu3572_task Schedule (maximum stream of network streams)

Problem Solving Report

Test instructions

Factory has m machine, need to do n tasks. For a task I. You need to spend a machine pi days, and the time to start doing this task is >=si, and the time to complete this task is <=ei.

For a task, it can only be completed by a machine. A machine can only do one task at a time.

However, a task can be divided into several periods of time to complete. Ask if you can finish all the tasks.

Ideas:

The network flow is modeled, and the modeling method is:

Think of each day and every task as a point. From the source point to each task. Build a PI-capacity edge (indicating how many days the task needs to be completed).

Every task to every day , if you can do the task on this day, build a capacity of 1 side. At last. Build a side capacity m (which indicates the maximum number of M tasks per machine) to the meeting point each day.

#include <map> #include <queue> #include <vector> #include <cstdio> #include <cstring># Include <iostream> #define INF 99999999using namespace std;int n,m,l[2010],head[2010],cnt,m;struct node{int v,w,n ext;}    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;//does not add optimization will 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 ScheduleTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 3311 Accepted Submission (s): 1154


Problem Descriptionour Geometry Princess XMM has stoped she study in computational geometry to concentrate on her newly op Ened Factory. Her factory had introduced M new machines in order to process the coming N tasks. For the i-th task, the factory have to start processing it on or after day Si, process it for Pi days, and finish the task Before or at day Ei. A machine can have work on one task at a time, and each task can be processed by at a time. However, a task can is interrupted and processed on different machines on different days.
Now she wonders whether he had a feasible schedule to finish all the tasks in time. She turns to the help.

Inputon The first line comes an integer T (t<=20), indicating the number of test cases.

You are given-N (n<=500) and M (m<=200) on the first line of all test case. Then on each of next N lines is three integers Pi, si and Ei (1<=pi, Si, ei<=500), which has the meaning described In the description. It is guaranteed, this in a feasible schedule every task, can be finished would be do before or at its end day.

Outputfor 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

24 31 3 5 1 1 42 3 73 5 92 22 1 31 2 2

Sample Output

Case 1:yes Case   2:yes

Authorallenlowesy
Source

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 (--host by UESTC)

Hdu3572_task Schedule (maximum stream of network streams)