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:
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
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.
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.
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 1 2 3 2 4
Hint
Explanation of the sample:
Here‘s a graphical schedule for this output:
Here‘s a graphical schedule for this output:
Time 1 2 3 4 5 6 7 8 9 10Other outputs using the same number of stalls are possible.
Stall 1 c1>>>>>>>>>>>>>>>>>>>>>>>>>>>
Stall 2 .. c2>>>>>> c4>>>>>>>>> .. ..
Stall 3 .. .. c3>>>>>>>>> .. .. .. ..
Stall 4 .. .. .. c5>>>>>>>>> .. .. ..
思路：
首先根据挤奶时间的先后顺序排序。。。然后将第一头牛加入优先队列。。然后就是加入优先队列的牛应该根据越早结束挤奶那么优先级更高，如果时间结束点相等，那么开始时间早的优先级高。。。
然后从前向后枚举。如果碰到有牛的挤奶时间的开始值大于优先队列的首部的结束值，那么说明这两头牛可以一起公用一个挤奶房。。然后从优先队列中删除这头牛。。那么这个问题就得到解决了。。。这道题用优先队列主要是又换时间；
#include<string.h> #include<stdio.h> #include<stack> #include<queue> #include<vector> #include<algorithm> using namespace std; struct node { int start_time; int end_time; int num; friend bool operator<(node a,node b) { if( a.end_time==b.end_time) return a.start_time>b.start_time; return a.end_time>b.end_time; } }arr[50050],now; bool cmp(node a,node b) { if(a.start_time==b.start_time) return a.end_time<b.end_time; return a.start_time<b.start_time; } int ans[50050]={0}; int main() { int i,n,l=1,MAX=-1; while(scanf("%d",&n)!=-1) { l=1; priority_queue<node>s,q; for(i=0;i<n;i++) { scanf("%d%d",&arr[i].start_time,&arr[i].end_time); arr[i].num=i; } sort(arr,arr+n,cmp); s.push(arr[0]); ans[s.top().num]=l; for(i=1;i<n;i++) { now=s.top(); if(arr[i].start_time>now.end_time) { ans[arr[i].num]=ans[now.num]; s.pop(); } else ans[arr[i].num]=++l; s.push(arr[i]); } printf("%d\n",l); for(i=0;i<n;i++) { printf("%d\n",ans[i]); } } return 0; }
