POJ 3190 Stall Reservations

Stall Reservations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15069		Accepted: 5270		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.

 

题意：一头牛在同一时间只能处理一个区间，问处理所有区间最少需要几个区间，并输出牛的安排情况

 

题解：将所有区间[l，r]排序，l不相等时小l排在前面，相等时小r排在前面，再用优先队列q保存每头牛的工作区间，越快结束的放在前面（r不相等时，小r放前面，r相等时，大l放前面）,按顺序处理结构体p里面的牛，能安排队列q中的牛就优先安排，安排不了就加牛

 

#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
#include<queue>
#include<set>
#define ll long long
using namespace std;
struct node
{
    int id;
    int l;
    int r;
}p[1000005],temp,now;

int vis[1000005];
bool cmp(node x,node y)//将开始处理时间早，处理时间短的放前面
{
    if(x.l!=y.l)
        return x.l<y.l;
    else
        return x.r<y.r;
}
bool operator < (node x,node y)//二次处理的时候，将结束时间早，开始时间晚的放前面
{
    if(x.r!=y.r)
        return x.r>y.r;
    else
        return x.l<y.l;
}

priority_queue<node>q;
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&p[i].l,&p[i].r);
            p[i].id=i;
        }
        sort(p,p+n,cmp);
        int cnt=1;
        q.push(p[0]);
        vis[p[0].id]=1;
        for(int i=1;i<n;i++)
        {
            if(!q.empty()&&q.top().r<p[i].l)
            {
                vis[p[i].id]=vis[q.top().id];
                q.pop();
            }
            else
            {
                cnt++;
                vis[p[i].id]=cnt;
            }
            q.push(p[i]);
        }
        printf("%d
",cnt);
        for(int i=0;i<n;i++)
            printf("%d
",vis[i]);
        while(!q.empty())
            q.pop();
    }
    return 0;
}