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

2016-04-11 · 2 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
************************************************ */

 1#include <cstdio>
 2#include <cstring>
 3#include <iostream>
 4#include <algorithm>
 5#include <vector>
 6#include <queue>
 7#include <set>
 8#include <map>
 9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair

 1using namespace std;
 2const double eps = 1E-8;
 3const int dx4[4]={1,0,0,-1};
 4const int dy4[4]={0,-1,1,0};
 5const int inf = 0x3f3f3f3f;
 6const int N=1E6+7;
 7int n;
 8int p[N];
 9int main()
10{
11	#ifndef  ONLINE_JUDGE 
12	freopen("code/in.txt","r",stdin);
13  #endif
14	scanf("%d",&n);
15	ms(p,0);
16	for ( int i = 1 ; i <= n ; i++)
17	{
18	    int l,r;
19	    scanf("%d %d",&l,&r);
20	    p[l]++;
21	    p[r+1]--;
22	}
23	for ( int i = 1 ; i <= 1000000 ; i++)
24	{
25	    p[i] = p[i]+p[i-1];
26	}

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

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

1  #ifndef ONLINE_JUDGE  
2  fclose(stdin);
3  #endif
4    return 0;
5}