Stall Reservations(POJ 3190 贪心+优先队列)

Stall Reservations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4434		Accepted: 1588		Special Judge

Description

Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over some precise time interval A..B (1 <= A <= B <= 1,000,000), which includes both times A and B. Obviously, FJ must create a reservation system to determine which stall each cow can be assigned for her milking time. Of course, no cow will share such a private moment with other cows.

Help FJ by determining:

The minimum number of stalls required in the barn so that each cow can have her private milking period

An assignment of cows to these stalls over time

Many answers are correct for each test dataset; a program will grade your answer.
Input

Line 1: A single integer, N

Lines 2..N+1: Line i+1 describes cow i's milking interval with two space-separated integers.
Output

Line 1: The minimum number of stalls the barn must have.

Lines 2..N+1: Line i+1 describes the stall to which cow i will be assigned for her milking period.
Sample Input

5
1 10
2 4
3 6
5 8
4 7

Sample Output

4
1
2
3
2
4

Hint

Explanation of the sample:

Here's a graphical schedule for this output:

Time     1  2  3  4  5  6  7  8  9 10

Stall 1 c1>>>>>>>>>>>>>>>>>>>>>>>>>>>

Stall 2 .. c2>>>>>> c4>>>>>>>>> .. ..

Stall 3 .. .. c3>>>>>>>>> .. .. .. ..

Stall 4 .. .. .. c5>>>>>>>>> .. .. ..

Other outputs using the same number of stalls are possible.

这个题是说一些奶牛要在指定的时间内挤牛奶，而一个机器只能同时对一个奶牛工作。给你每头奶牛的指定时间的区间，问你最小需要多少机器。

先按奶牛要求的时间起始点进行从小到大排序，然后维护一个优先队列，里面以已经开始挤奶的奶牛的结束时间早为优先。然后每次只需要检查当前是否有奶牛的挤奶工作已经完成的机器即可，若有，则换那台机器进行工作。若没有，则加一台新的机器。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn=60000;
int n,use[maxn];
struct Node
{
int l;
int r;
int pos;
bool operator <(const Node &a)const
{
if(r==a.r)
return l>a.l;
return r>a.r;
}
}a[maxn];
priority_queue<Node> q;
bool cmp(Node a,Node b)
{
if(a.l==b.l)
return a.r<b.r;
return a.l<b.l;
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
for(int i=0;i<n;i++)
{
scanf("%d%d",&a[i].l,&a[i].r);
a[i].pos=i;
}
sort(a,a+n,cmp);
q.push(a[0]);
int now=0,ans=1;
use[a[0].pos]=1;
for(int i=1;i<n;i++)
{
if(!q.empty()&&q.top().r<a[i].l)
{
use[a[i].pos]=use[q.top().pos];
q.pop();
}
else
{
ans++;
use[a[i].pos]=ans;
}
q.push(a[i]);
}
printf("%d\n",ans);
for(int i=0;i<n;i++)
printf("%d\n",use[i]);
while(!q.empty())
q.pop();
}
return 0;
}

TLE CODE:

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
struct node
{
int x1,x2;
int num;
}a[50000+10];
bool cmp(node a,node b)
{
if(a.x1==b.x1)
return a.x2<b.x2;
return a.x1<b.x1;
}
bool vis[50000+10];
int pos[50000+10];
int n;
int search(int m)
{
int l=0,r=n-1,mid,k=-1;
while(l<=r)
{
mid=(l+r)/2;
if(a[mid].x1>=m)
{

k=mid;
r=mid-1;
}
else
l=mid+1;
}
return k;
}
int main()
{
int i,j;
freopen("in.txt","r",stdin);
while(scanf("%d",&n)!=EOF)
{
int count=0;
memset(vis,0,sizeof(vis));
fill(pos,pos+n,0);
for(i=0;i<n;i++)
{
scanf("%d%d",&a[i].x1,&a[i].x2);
a[i].num=i;
}
sort(a,a+n,cmp);
int last,coun=0,p;
for(i=0;i<n;i++)
{
if(vis[i])
continue;
last=a[i].x2+1;
vis[i]=1;
pos[a[i].num]=++coun;
while(1)
{
p=search(last);
if(p==-1)
break;
if(vis[p])
last=a[p].x1+1;
else
{
last=a[p].x2+1;
vis[p]=1;
pos[a[p].num]=coun;
}
}
}
printf("%d\n",coun);
for(i=0;i<n;i++)
printf("%d\n",pos[i]);
}
}