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

Sample Output

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;
}