POJ 3190 Stall Reservations

Stall Reservations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions:8581		Accepted:2981		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.

根据开始时间排序，然后每到一头牛时判断前面是否有空出的
位置，若有则放入，没有则加1
代码：

#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
struct node
{
    int num,st,en;
}cow[50004];
bool cmp1(node a,node b)
{
    return a.st<b.st;
}
struct node2
{
    int dec,time;
};
bool operator<(node2 a,node2 b)
{
    return a.time>b.time;
}
priority_queue<node2>P;
struct node3
{
    int dec,num;
}c[50003];
bool cmp2(node3 a,node3 b)
{
    return a.num<b.num;
}
int main()
{
    int n;scanf("%d",&n);
    int i,j;
    for(i=1;i<=n;i++)
    {
        scanf("%d%d",&cow[i].st,&cow[i].en);
        cow[i].num=i;
    }
    sort(cow+1,cow+n+1,cmp1);
    int ans=0;
    for(i=1;i<=n;i++)
    {
        int x=cow[i].st,y=cow[i].en,num=cow[i].num;
        if(P.empty()||P.top().time>=x)
        {
            ans++;
            c[i].dec=ans;c[i].num=num;
            node2 e;e.dec=ans;e.time=y;
            P.push(e);
        }
        else
        {
            node2 e=P.top();P.pop();
            c[i].dec=e.dec;c[i].num=num;
            e.time=y;
            P.push(e);
        }
    }
    sort(c+1,c+n+1,cmp2);
    printf("%d\n",ans);
    for(i=1;i<=n;i++)
        printf("%d\n",c[i].dec);
    return 0;
}