洛谷P2859 [USACO06FEB]摊位预订Stall Reservations

P2859 [USACO06FEB]摊位预订Stall Reservations

题目描述

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 periodAn assignment of cows to these stalls over timeMany answers are correct for each test dataset; a program will grade your answer.

约翰的N(l<N< 50000)头奶牛实在是太难伺候了，她们甚至有自己独特的产奶时段.当 然对于某一头奶牛，她每天的产奶时段是固定的，为时间段A到B包括时间段A和时间段B.显然，约翰必须开发一个调控系统来决定每头奶牛应该被安排到哪个牛 棚去挤 奶，因为奶牛们显然不希望在挤奶时被其它奶牛看见.

约翰希望你帮他计算一下：如果要满足奶牛们的要求，并且每天每头奶牛都要被挤过奶，至少需要多少牛棚 •每头牛应该在哪个牛棚被挤奶。如果有多种答案，你只需任意一种即可。

输入输出格式

输入格式：

Line 1: A single integer, N

Lines 2..N+1: Line i+1 describes cow i's milking interval with two space-separated integers.

输出格式：

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.

输入输出样例

输入样例#1：

5
1 10
2 4
3 6
5 8
4 7

输出样例#1：

4
1
2
3
2
4

说明

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.

由@zhouyonglong提供spj

贪心。先按找开始时间排序。然后把第一个开始的放到堆里，堆里的元素按照结束时间先后排序，结束早的排在前面。

然后从第二个开始，逐个扫，每一个与堆顶元素比，如果开始时间比对顶元素的结束时间晚，那么两个就可以共用

一个牛棚。

#include <bits/stdc++.h>
inline void read(int &x){x = 0;char ch = getchar();char c = ch;while(ch > '9' && ch < '0')c = ch, ch = getchar();while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();if(c == '-')x = -x;}
inline int max(int &a, int &b){return a > b ? a : b;}
inline int min(int &a, int &b){return a > b ? b : a;}
inline void swap(int &a, int &b){int tmp = a;a = b;b = tmp;}
const int INF = 0x3f3f3f3f;
const int MAXN = 50000 + 10;
const int MAXNUM = 1000000 + 10;

int s[MAXN],t[MAXN],rank[MAXN],node[MAXN];

bool cmp1(int a,int b)
{
	return s[a] <= s[b];
}

struct cmp2
{
	bool operator()(int a,int b)
	{
		return t[a] >= t[b];
	}
};

std::priority_queue<int, std::vector<int>, cmp2> q;

int main()
{
	register int cnt = 0;
	register int n;
	read(n);
	for(int i = 1;i <= n;i ++)
	{
		read(s[i]);read(t[i]);
		rank[i] = i;
	}
	std::sort(rank + 1, rank + 1 + n, cmp1);
	q.push(rank[1]);
	node[rank[1]] = ++cnt;
	for(register int i = 2;i <= n;i ++)
	{
		register int now = rank[i];
		register int top = q.top();
		if(s[now] > t[top])
		{
			node[now] = node[top];
			q.pop();
			q.push(rank[i]);
		}
		else
		{
			node[now] = ++cnt;
			q.push(now);
		}
	}
	printf("%d\n", cnt);
	for(int i = 1;i <= n;i ++)
		printf("%d\n", node[i]);
	return 0;
}