Stall Reservations

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.

 

poj链接： http://poj.org/problem?id=3190

 

 

代码：

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