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.

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

题意：有n项任务和m台机器，给出每个任务的执行时间pi，和开始时间si，以及截止时间ei。每项任务需要在si或si后开始执行，在ei或ei前执行完毕，每个任务可以分段，在任何一台机器执行，每台机器在一个时间只能执行一个任务。。

题解：拆点的最大流判断是否满流，建图是重点。考虑每个任务可以在任何一台机器执行，因此每个任务在规定的时间内就可以分到任何一台机器（一条机器在一个时间点只执行一个任务），所以将时间进行拆点，添加一个源点S，从源点S向每个任务连边，因为每个任务的执行时间是pi，所以从超级源点连向每个任务的流是pi，即建边add_edge (S,i,pi)；然后考虑每个任务的执行，一个任务只能同时在一个时间点被工作，即不能同时既在时间点A上加工又在时间点B上加工，,所以每一个任务 i 向时间点[s,e]连一条容量为1的边; add_edge (i,s--->t,1)；最后每一个时间点向T连一条容量为m个边，说明一个时间点只能最多同时有m个机器在工作。最后判断是否漫满流，即判断从S流出的n?p个流量能否全部流入T中。

参考代码：

#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;
}