#POJ3190#Stall Reservations（贪心 -> 冲突分配）

Stall Reservations

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7008 Accepted: 2519 Special Judge

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

Many answers are correct for each test dataset; a program will grade your answer.

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

51 102 43 65 84 7

Sample Output

412324

Hint

Explanation of the sample: 

Here's a graphical schedule for this output: 

Time     1  2  3  4  5  6  7  8  9 10Stall 1 c1>>>>>>>>>>>>>>>>>>>>>>>>>>>Stall 2 .. c2>>>>>> c4>>>>>>>>> .. ..Stall 3 .. .. c3>>>>>>>>> .. .. .. ..Stall 4 .. .. .. c5>>>>>>>>> .. .. ..

Other outputs using the same number of stalls are possible.

题意：

有N头奶牛，（1<=N<=50000）,每头奶牛只在特定的时间段内[A,B]内才愿意挤奶(1<=A,B<=1000000)。

一个挤奶机只能同时挤一头奶牛。农夫约翰最少需要准备多少个挤奶机，才能满足所有奶牛的挤奶需求。

这题的贪心其实很容易会想到要去找冲突矛盾最多的就是机器数，

（整体按左端点从小到大排序）

然后问题就是每只奶牛的分配了，优先队列的用处就在于将右端点从小到大排序，

以保证最先空出来的尽量能够被利用

我的朋友还有另一种解法，（虽然据说是在午睡梦中想出来的），但是很显然也是可行的，

将端点全部单独取出，标记是左还是右，从小到大排序（相同时左端点在前），

然后用普通队列来存放空出的机器，之后扫一遍，

其实原理差不多，都保证了机器能被最大化利用，等他有空了我去扒扒他的代码再发过来

（嗯我是奸商）

Code:

StatusAcceptedTime391msMemory2036kBLength1473LangG++Submitted2017-07-14 09:43:40Shared

#include<iostream> #include<cstdio> #include<cstdlib> #include<cmath> #include<algorithm> #include<vector> #include<queue> #include<cstring> using namespace std;   const int Max = 50000;   struct node{     int l, r; int pos;    bool operator < (const node & X) const{         if(l == X. l)   return r < X.r;         return l < X.l;     } bool operator > (const node & X) const{//if(r == X.r)return l > X.l;return r > X. r;}}Cow[Max + 5];   int N; int Room[Max + 5];priority_queue<node, vector<node>, greater<node> >Q;  bool  getint(int & num){     char c;    int flg = 1;    num = 0;     while((c = getchar()) < '0' || c > '9'){         if(c == '-')    flg = -1;         if(c == -1)     return 0;     }     while(c >= '0' && c <= '9'){             num = num * 10 + c - 48;         if((c = getchar()) == -1)   return 0;     }     num *= flg;     return 1; }   int main(){ while(getint(N)){for(int i = 1; i <= N; ++ i)getint(Cow[i].l), getint(Cow[i].r), Cow[i].pos = i;sort(Cow + 1, Cow + 1 + N);int R = 1;Q.push(Cow[1]);Room[Cow[1].pos] = R;for(int i = 2; i <= N; ++ i){if(! Q.empty() && Q.top().r < Cow[i].l){Room[Cow[i].pos] = Room[Q.top().pos];Q.pop();Q.push(Cow[i]);}else {++ R;Room[Cow[i].pos] = R;Q.push(Cow[i]);}}printf("%d\n", R);for(int i = 1; i <= N; ++ i)printf("%d\n", Room[i]);while(! Q.empty())Q.pop();}    return 0; }