POJ 3190 Stall Reservations

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.

 

 一群奶牛，只在固定的时间产奶，每头牛只在固定的时间产奶，每头牛都需要一个挤奶的机器

 所有奶牛产奶，最少需要多少个机器，输出每一个奶牛使用挤奶机器的编号

 

 

  这题贪心 ，需要用优先队列优化 ， 

 

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
const int maxn=50010;
struct node {
    int x,y,id;
    friend bool operator<(node a,node b) {
        return a.y>b.y;
    }
} qu[maxn];
int cmp(node a,node b) {
    return a.x<b.x;
}
int ID[maxn];
int main() {
    int n;
    while(scanf("%d",&n)!=EOF ) {
        for (int i=0 ; i<n ; i++) {
            scanf("%d%d",&qu[i].x,&qu[i].y);
            qu[i].id=i;
        }
        sort(qu,qu+n,cmp);
        memset(ID,0,sizeof(ID));
        priority_queue<node>q;
        int ans=1;
        q.push(qu[0]);
        ID[qu[0].id]=1;
        for (int i=1 ; i<n ; i++) {
            if (q.top().y<qu[i].x) {
                ID[qu[i].id]=ID[q.top().id];
                q.pop();
                q.push(qu[i]);
            } else {
                ID[qu[i].id]=++ans;
                q.push(qu[i]);
            }
        }
        printf("%d
",ans);
        for (int i=0 ; i<n ; i++) {
            printf("%d
",ID[i]);
        }
    }
    return 0;
}