Task Schedule


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6422    Accepted Submission(s): 2024


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


 



Author


allenlowesy


 



Source


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


 


题意：工厂有m台机器，需要做n个任务。对于一个任务i，你需要花费一个机器Pi天，而且，开始做这个任务的时间要>=Si，完成这个任务的时间<=Ei。对于一个任务，只能由一个机器来完成，一个机器同一时间只能做一个任务。但是，一个任务可以分成几段不连续的时间来完成。问，能否做完全部任务。

诈一看，感觉是一道贪心题……这种题用网络流做的话，难都不在于网络流算法，而是建图的过程，因为这是一个很抽象的过程……这里我们建图的过程就是：把每一天和每个任务看做点。由源点到每一任务，建容量为pi的边（表示任务需要多少天完成）。每个任务到每一天，若是可以在这天做任务，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。


#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <algorithm>
using namespace std;
const int maxn = 1111;
const int maxm = 505000;
const int INF = 0xfffffff;
int idx;
int cur[maxn],pre[maxn],dis[maxn],gap[maxn],aug[maxn],head[maxn];
struct Node
{
  int u,v,w,next;
}edge[maxm];
void addEdge(int u,int v,int w)
{
  edge[idx].u = u;
  edge[idx].v = v;
  edge[idx].w = w;
  edge[idx].next = head[u];
  head[u] = idx++;
  
  edge[idx].u = v;
  edge[idx].v = u;
  edge[idx].w = 0;
  edge[idx].next = head[v];
  head[v] = idx++;
}
int SAP(int s,int e,int n)
{
  int max_flow = 0,v,u = s;
  int id,mindis;
  aug[s] = INF;
  pre[s] = -1;
  memset(dis,0,sizeof(dis));
  memset(gap,0,sizeof(gap));
  gap[0] = n;
  for(int i=0; i<=n; i++)
    cur[i] = head[i];// 初始化当前弧为第一条弧
  while(dis[s] < n)
  {
    bool flag = false;
    if(u == e)
    {
      max_flow += aug[e];
      for(v=pre[e]; v!=-1; v=pre[v])// 路径回溯更新残留网络
      {
        id = cur[v];
        edge[id].w -= aug[e];
        edge[id^1].w += aug[e];
        aug[v] -= aug[e];// 修改可增广量，以后会用到
        if(edge[id].w == 0) u = v;// 不回退到源点，仅回退到容量为0的弧的弧尾
      }
    }
    for(id=cur[u]; id!=-1; id=edge[id].next)
    {// 从当前弧开始查找允许弧
      v = edge[id].v;
      if(edge[id].w>0 && dis[u]==dis[v]+1) // 找到允许弧
      {
        flag = true;
        pre[v] = u;
        cur[u] = id;
        aug[v] = min(aug[u], edge[id].w);
        u = v;
        break;
      }
    }
    if(flag == false)
    {
      if(--gap[dis[u]] == 0) break;/* gap优化，层次树出现断层则结束算法 */
      mindis = n;
      cur[u] = head[u];
      for(id=head[u]; id!=-1; id=edge[id].next)
      {
        v = edge[id].v;
        if(edge[id].w>0 &&dis[v]<mindis)
        {
          mindis = dis[v];
          cur[u] = id;// 修改标号的同时修改当前弧
        }
      }
      dis[u] = mindis+1;
      gap[dis[u]]++;
      if(u != s) u = pre[u];// 回溯继续寻找允许弧
    }
  }
  return max_flow;
}
int main()
{
  int t,n,m,pi,si,ei;
  int Max,sum,source,sink,vn;
  scanf("%d",&t);
  for(int cas=1; cas<=t; cas++)
  {
    idx = 0;
    memset(head,-1,sizeof(head));
    sum = 0,source = 0,Max = 0;
    scanf("%d%d",&n,&m);
    for(int i=1; i<=n; i++)
    {
      scanf("%d%d%d",&pi,&si,&ei);
      sum += pi;
      Max = max(Max,ei);
      addEdge(source,i,pi);
      for(int j=si; j<=ei; j++)
        addEdge(i,n+j,1);
    }
    sink = n+Max+1;
    vn = sink+1;
    for(int i=1; i<=Max; i++)
    {
      addEdge(n+i,sink,m);
    }
    if(SAP(source,sink,vn) == sum)
      printf("Case %d: Yes\n\n",cas);
    else printf("Case %d: No\n\n",cas);
  }
}