HDU3572_Task Schedule(网络流最大流)

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
");
        else printf("No
");
        printf("
");
    }
    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

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

Sample Output

Case 1: Yes
   
Case 2: Yes

Author

allenlowesy

Source

2010 ACM-ICPC Multi-University Training Contest（13）——Host by UESTC