BZOJ 1651: [Usaco2006 Feb]Stall Reservations 专用牛棚 (前缀和)

Apr 11, 2016 · 3 min read · 前缀和

1651: [Usaco2006 Feb]Stall Reservations 专用牛棚

Time Limit: 10 Sec  Memory Limit: 64 MB Submit: 700  Solved: 393 [Submit][Status][Discuss]

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

有N头牛,每头牛有个喝水时间,这段时间它将专用一个Stall 现在给出每头牛的喝水时间段,问至少要多少个Stall才能满足它们的要求

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

OUTPUT DETAILS:

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.

HINT

不妨试下这个数据,对于按结束点SORT,再GREEDY的做法 1 3 5 7 6 9 10 11 8 12 4 13 正确的输出应该是3

思路:求所有点中的最大的厚度就是答案。前缀和搞之。1A.

/* ***********************************************
Author :111qqz
Created Time :2016年04月11日 星期一 20时15分13秒
File Name :code/bzoj/1651.cpp
************************************************ */

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair

using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=1E6+7;
int n;
int p[N];
int main()
{
	#ifndef  ONLINE_JUDGE 
	freopen("code/in.txt","r",stdin);
  #endif
	scanf("%d",&n);
	ms(p,0);
	for ( int i = 1 ; i <= n ; i++)
	{
	    int l,r;
	    scanf("%d %d",&l,&r);
	    p[l]++;
	    p[r+1]--;
	}
	for ( int i = 1 ; i <= 1000000 ; i++)
	{
	    p[i] = p[i]+p[i-1];
	}

	int ans = 0 ;
	for ( int i = 1 ; i <= 1000000 ; i++) ans = max(ans,p[i]);

	printf("%d\n",ans);

  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}