POJ：3190-Stall Reservations

Stall Reservations

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 9173 Accepted: 3208 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.

解题心得：

题意就是每个牛在一段时间内要产奶，每个奶牛在产奶的时候必须给它安排一间屋子，问最少需要多少间屋子，以及给每个奶牛安排产奶的屋子是第几间。
也就是一个思维题，先把奶牛用起始时间拍个序，用一个优先队列1来放所有的屋子编号，再用一个优先队列2来记录每个奶牛产奶结束的时间，遇到一个奶牛开始产奶，就看优先队列2中在当前时间是否有奶牛已经结束产奶了，如果已经结束产奶，就将安排给那个产奶的奶牛屋子还回队列1当中，然后从队列1中找一个编号最小的屋子给当前的奶牛。优先队列1中出现的房间号数字最大的就是最少需要安排的房间数。

#include <stdio.h>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn = 5e4+100;
struct COW {
    int s,e,num;
    bool operator <(const COW & a) const {
        return a.e < e;
    }
}cow[maxn];
int n;

bool cmp(COW a,COW b) {
    if(a.s  == b.s)
        return a.e < b.e;
    return a.s < b.s;
}

void init() {
    for(int i=1;i<=n;i++) {
        scanf("%d%d",&cow[i].s,&cow[i].e);
        cow[i].num = i;
    }
    sort(cow+1,cow+n+1,cmp);
}

int ans[maxn];
void get_ans() {
    int Max_num = -1;
    priority_queue <int,vector<int>,greater <int> > qu;
    priority_queue <COW> qu2;
    for(int i=1;i<=maxn;i++)
        qu.push(i);
    for(int i=1;i<=n;i++) {
        COW now = cow[i];
        while(!qu2.empty() && qu2.top().e < now.s) {
            qu.push(ans[qu2.top().num]);
            qu2.pop();
        }
        ans[now.num] = qu.top();
        if(qu.top() > Max_num)
            Max_num = qu.top();
        qu.pop();
        qu2.push(cow[i]);
    }
    printf("%d
",Max_num);
    for(int i=1;i<=n;i++)
        printf("%d
",ans[i]);
}

int main() {
    scanf("%d",&n);
    init();
    get_ans();
    return 0;
}