POJ-3190 Stall Reservations (优先队列+贪心）

题目：

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

---

思路：

• 一开始以为这题跟我以前做的一道求最大重叠区间数一样，然后发现要求每头牛安排在哪个棚就不能按那题的方法；
• 我用当时做的比较笨的方法一个个棚求可以合并的牛，过了样例，但由于区间范围比那题大很多所以不出所料TLE；
• 一开始想用离散化优化没写出来，后来看了别人的博客才知道用优先队列优化更方便。

Code：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>

using namespace std;
const int MAXN = 50010;
int cnt[MAXN];

struct node {
    int l, r, id;
    friend bool operator < (node a, node b) {
        return a.r > b.r;
    }
}a[MAXN];

bool cmp(node a, node b) {
    if (a.l == b.l) return a.r < b.r;
    return a.l < b.l;
}

priority_queue<node> q;

int main() {
    std::ios::sync_with_stdio(false);
    int n;
    while(cin>>n) {
        for (int i = 0; i < n; ++i) {
            cin>>a[i].l>>a[i].r;
            a[i].id = i;
        }
        sort(a, a+n, cmp);
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (!q.empty() && q.top().r < a[i].l) {
                cnt[a[i].id] = cnt[q.top().id];
                q.pop();
            }
            else cnt[a[i].id] = ++ans;
            q.push(a[i]);
        }
        cout<<ans<<endl;
        for (int i = 0; i < n; ++i) cout<<cnt[i]<<endl;
    }

    return 0;
}