Stall Reservations - POJ 3190 贪心

Stall Reservations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3063		Accepted: 1104		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：每个奶牛有固定挤奶的时间段，每个房间最多同时容纳一只奶牛，问至少需要多少房间，以及每只奶牛去哪个房间。

思路：挑战书上贪心部分的题，但我没看出来有多贪心→_→，大概就是先离散化一下数据（但我不清楚不离散化是不是会超时），然后奶牛分别按开始时间和结束时间排序，队列qu中记录还有哪些是空的房间，然后模拟即可。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
struct node
{
    int l,r,pos;
}cow1[50010];
bool cmp(node a,node b)
{
    return a.l<b.l;
}
struct node2
{
    int l,r,pos;
}cow2[50010];
bool cmp2(node2 a,node2 b)
{
    return a.r<b.r;
}
int n,m,num1,num2,point[1000010],f[100010];
queue<int> qu;
int main()
{
    int i,j,k,maxn;
    while(~scanf("%d",&n))
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d%d",&cow1[i].l,&cow1[i].r);
            f[i*2-1]=cow1[i].l;
            f[i*2]=cow1[i].r;
        }
        sort(f+1,f+1+n*2);
        m=0;maxn=0;
        for(i=1;i<=n*2;i++)
           if(f[i]!=f[i-1])
           {
               m++;
               point[f[i]]=m;
           }
        for(i=1;i<=n;i++)
        {
            cow2[i].l=cow1[i].l=point[cow1[i].l];
            cow2[i].r=cow1[i].r=point[cow1[i].r];
            cow2[i].pos=cow1[i].pos=i;
        }
        sort(cow1+1,cow1+1+n,cmp);
        sort(cow2+1,cow2+1+n,cmp2);
        while(!qu.empty())
           qu.pop();
        num1=1;num2=1;
        for(i=1;i<=m;i++)
        {
            while(cow1[num1].l==i && num1<=n)
            {
                if(qu.empty())
                {
                   maxn++;
                   f[cow1[num1].pos]=maxn;
                }
                else
                {
                    k=qu.front();
                    f[cow1[num1].pos]=k;
                    qu.pop();
                }
                num1++;
            }
            while(cow2[num2].r==i && num2<=n)
            {
                qu.push(f[cow2[num2].pos]);
                num2++;
            }
        }
        printf("%d\n",maxn);
        for(i=1;i<=n;i++)
           printf("%d\n",f[i]);
    }
}