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POJ-3190-贪心（优先队列）

2014-07-15 23:36 519 查看

Stall Reservations

Time Limit: 1000MSMemory Limit: 65536K
Total Submissions: 2778Accepted: 990Special 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.
Source
USACO 2006 February Silver

该题的解题思路就不多说了，优先队列。看代码应该能够很好地理解！

#include<stdio.h>
#include<queue>
#include<algorithm>
using namespace std;

struct P
{
    int x,y;
    int p;
    friend bool operator<(P m,P n)
    {
        return m.y>n.y;  //右边界最小值优先
    }
}a[50005];

bool cmp(P m,P n)
{
    return m.x<n.x;
}

int main()
{
    int n;
    int b[50005];
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d%d",&a[i].x,&a[i].y);
        a[i].p=i;
    }
    sort(a,a+n,cmp);
    int z=0;
    b[a[0].p]=++z;
    priority_queue<P>q;
    q.push(a[0]);
    for(int i=1;i<n;i++)
    {
        P t=q.top();
        if(a[i].x>t.y)
        {
            b[a[i].p]=b[t.p];
            q.pop();
            q.push(a[i]);
        }
        else
        {
            b[a[i].p]=++z;
            q.push(a[i]);
        }
    }
    printf("%d\n",z);
    for(int i=0;i<n;i++)
        printf("%d\n",b[i]);
    return 0;
}


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