POJ3190 Stall Resevrations

POJ3190 Stall Resevrations

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.

  • 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

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1 10
2 4
3 6
5 8
4 7

Sample Output

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1
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2
4

Hint

Explanation of the sample:

Other outputs using the same number of stalls are possible.

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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.

Answer

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#include <cstdio>
#include <algorithm>
#include <queue>

using namespace std;

struct Cow{
int first, second;
int id;
bool operator< (const Cow &c) const{
return first < c.first;
}
}cows[50010];

int ans[50010];

priority_queue <pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > pq;

int main() {
int n, i, j, cnt = 0;

scanf("%d", &n);
for(i = 0; i < n; ++i) {
scanf("%d%d", &cows[i].first, &cows[i].second);
cows[i].id = i;
}

sort(cows, cows + n);
for(i = 0; i < n; ++i) {
if(pq.empty() || cows[i].first <= pq.top().first) {
ans[cows[i].id] = ++cnt;
pq.push(make_pair(cows[i].second, cnt));
} else {
ans[cows[i].id] = pq.top().second;
pq.pop();
pq.push(make_pair(cows[i].second, ans[cows[i].id]));
}
}

printf("%d\n", cnt);
for(i = 0; i < n; ++i) {
printf("%d\n", ans[i]);
}
return 0;
}

Analysis

  1. 每头奶牛有自己的产奶时间区间,现在要建一批牛栏,使得每个牛都能独自在牛栏里产奶,求最少牛栏数和牛与牛栏的分配关系。
  2. 将牛按开始时间排序,最小堆维护每个牛栏最右端点的值,对每个牛,若其开始时间大于堆顶,则放入此牛栏,否则新建牛栏。
# POJ

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