POJ

Stall Reservations

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7728 Accepted: 2736 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：一共n头牛 给出每头牛挤奶的时间 问最少要开几个挤奶的地方

#include <iostream>#include <cstdio>#include <queue>#include <algorithm>using namespace std;struct z{int x;int y;int num;bool operator < (const z &a)const{if(y==a.y)return a.x<x;return a.y<y;}};bool comp(const z &a,const z &b){if(a.x==b.x)return a.y<b.y;return a.x<b.x;}int main(){int n,i;while(scanf("%d",&n)!=EOF){struct z cow[n+1];int order[n+1];priority_queue<z> q;for(i=1;i<=n;i++){int x,y;scanf("%d%d",&cow[i].x,&cow[i].y);cow[i].num=i;}sort(cow+1,cow+1+n,comp);int count=1;order[cow[1].num]=1;q.push(cow[1]);for(i=2;i<=n;i++){int flag=0;if(q.size()&&cow[i].x>q.top().y){flag=1;order[cow[i].num]=order[(q.top().num)];q.pop();}if(flag==0){count++;order[(cow[i].num)]=count;}q.push(cow[i]);}printf("%d\n",count);for(i=1;i<=n;i++)printf("%d\n",order[i]);}return 0;}