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.
代码实现
1/* ***********************************************
2Author :111qqz
3Created Time :2016年04月11日 星期一 20时15分13秒
4File Name :code/bzoj/1651.cpp
5************************************************ */
6
7#include <cstdio>
8#include <cstring>
9#include <iostream>
10#include <algorithm>
11#include <vector>
12#include <queue>
13#include <set>
14#include <map>
15#include <string>
16#include <cmath>
17#include <cstdlib>
18#include <ctime>
19#define fst first
20#define sec second
21#define lson l,m,rt<<1
22#define rson m+1,r,rt<<1|1
23#define ms(a,x) memset(a,x,sizeof(a))
24typedef long long LL;
25#define pi pair < int ,int >
26#define MP make_pair
27
28using namespace std;
29const double eps = 1E-8;
30const int dx4[4]={1,0,0,-1};
31const int dy4[4]={0,-1,1,0};
32const int inf = 0x3f3f3f3f;
33const int N=1E6+7;
34int n;
35int p[N];
36int main()
37{
38 #ifndef ONLINE_JUDGE
39 freopen("code/in.txt","r",stdin);
40 #endif
41 scanf("%d",&n);
42 ms(p,0);
43 for ( int i = 1 ; i <= n ; i++)
44 {
45 int l,r;
46 scanf("%d %d",&l,&r);
47 p[l]++;
48 p[r+1]--;
49 }
50 for ( int i = 1 ; i <= 1000000 ; i++)
51 {
52 p[i] = p[i]+p[i-1];
53 }
54
55 int ans = 0 ;
56 for ( int i = 1 ; i <= 1000000 ; i++) ans = max(ans,p[i]);
57
58 printf("%d\n",ans);
59
60 #ifndef ONLINE_JUDGE
61 fclose(stdin);
62 #endif
63 return 0;
64}