poj 3190 贪心+优先队列优化

Stall Reservations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4274		Accepted: 1530		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.

<span style="font-size:18px;color:#3366ff;">#include <iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
struct Node{
     int s,e,id,num;
     bool operator<(const Node &a) const
     {
         return a.e<e;
     }
}node[50005];
int cmp(Node a,Node b)
{
     if(a.s!=a.s)
        return a.e<b.e;
     else return a.s<b.s;
}
int cmp2(Node a,Node b)
{
    return a.id<b.id;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
            {
                scanf("%d %d",&node[i].s,&node[i].e);
                node[i].id=i;
            }
        sort(node+1,node+n+1,cmp);
        priority_queue<Node>  q;
        node[1].num=1;
        q.push(node[1]);
        int cnt=1;
        for(int i=2;i<=n;i++)
        {
             Node temp=q.top();
            if(node[i].s<=temp.e)
            {
                cnt++;
                node[i].num=cnt;
                q.push(node[i]);
            }
            else
            {
                node[i].num=temp.num;
                q.pop();
                q.push(node[i]);
            }
        }
        sort(node+1,node+n+1,cmp2);
        printf("%d\n",cnt);
        for(int i=1;i<=n;i++)
            printf("%d\n",node[i].num);
     }
    return 0;
}</span>