POJ2376 Cleaning Shifts

POJ2376 Cleaning Shifts

Description

Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleaning chores around the barn. He always wants to have one cow working on cleaning things up and has divided the day into T shifts (1 <= T <= 1,000,000), the first being shift 1 and the last being shift T.

Each cow is only available at some interval of times during the day for work on cleaning. Any cow that is selected for cleaning duty will work for the entirety of her interval.

Your job is to help Farmer John assign some cows to shifts so that (i) every shift has at least one cow assigned to it, and (ii) as few cows as possible are involved in cleaning. If it is not possible to assign a cow to each shift, print -1.

Input

* Line 1: Two space-separated integers: N and T

* Lines 2..N+1: Each line contains the start and end times of the interval during which a cow can work. A cow starts work at the start time and finishes after the end time.

Output

* Line 1: The minimum number of cows Farmer John needs to hire or -1 if it is not possible to assign a cow to each shift.

Sample Input

1
2
3
4
3 10
1 7
3 6
6 10

Sample Output

1
2

Hint

This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.

INPUT DETAILS:

There are 3 cows and 10 shifts. Cow #1 can work shifts 1..7, cow #2 can work shifts 3..6, and cow #3 can work shifts 6..10.

OUTPUT DETAILS:

By selecting cows #1 and #3, all shifts are covered. There is no way to cover all the shifts using fewer than 2 cows.

Answer

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#include <cstdio>
#include <algorithm>
using namespace std;

struct cow {
int s, t;
cow() {}
cow(int s, int t) : s(s), t(t) {}
};

bool cmp(cow a, cow b) {
return a.s < b.s;
}

int main() {
int N, T;
scanf("%d %d", &N, &T);
cow *a = new cow[N];

int x, y;
for (int i = 0; i < N; ++i) {
scanf("%d %d", &x, &y);
a[i] = cow(x, y);
}

sort(a, a+N, cmp);
int ans = 0, now = 1, i = 0;
while (now <= T) {
int t = -1;
for (; i < N && a[i].s <= now; ++i)
t = max(t, a[i].t);
if (t == -1) {
ans = -1;
break;
} else {
++ans;
now = t + 1;
}
}
printf("%d\n", ans);
return 0;
}

Analysis

  1. N 头牛,T 个连续时间段,选取最小的牛数完成 T 个时间段。
  2. 将牛按开始工作时间顺序排列,每次选取结束时间最大的牛(贪心),下一轮选取开始时间不大于上一轮结束时间+1,且结束时间最大的牛,直到 T 结束。
  3. 若结束时间小于 T,且没有更新,则安排失败。
# POJ

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