poj 3190 Stall Reservations（区间贪心，优先队列）

Stall Reservations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5390		Accepted: 1945		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.

Source

USACO 2006 February Silver

题意：每头牛独享一个牛栏里的[A, B]，问最少需要几个牛栏。

思路：优先选A最小的，维持一个牛栏B最小堆，将新来的奶牛塞进B最小的牛栏里。

代码：

#include
#include
#include
#include
using namespace std;
const int maxn = 50005;
struct node
{
    int x, y, id;
    bool operator <(const node a) const
    {
        return y > a.y;
    }
}a[maxn];
int ans[maxn];
bool cmp(node a, node b)
{
    if(a.x == b.x) return a.y < b.y;
    else return a.x < b.x;
}
int main(void)
{
    int n;
    while(~scanf("%d", &n))
    {
        int k = 1;
        priority_queue pq;
        for(int i = 0; i < n; i++)
        {
            scanf("%d%d", &a[i].x, &a[i].y);
            a[i].id = i;
        }
        sort(a, a+n, cmp);
        pq.push(a[0]);
        ans[a[0].id] = k++;
        for(int i = 1; i < n; i++)
        {
            node t = pq.top();
            if(t.y < a[i].x)
            {
                ans[a[i].id] = ans[t.id];
                pq.pop();
                pq.push(a[i]);
            }
            else
            {
                ans[a[i].id] = k++;
                pq.push(a[i]);
            }
        }
        printf("%d\n", k-1);
        for(int i = 0; i < n; i++)
            printf("%d\n", ans[i]);
    }
    return 0;
}