POJ 3190 Stall Reservations
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Stall Reservations
Time Limit: 1000MS | Memory Limit: 65536K | |||
Total Submissions: 15069 | Accepted: 5270 | 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:
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>>>>>>>>> .. .. ..
题意:一头牛在同一时间只能处理一个区间,问处理所有区间最少需要几个区间,并输出牛的安排情况
题解:将所有区间[l,r]排序,l不相等时小l排在前面,相等时小r排在前面,再用优先队列q保存每头牛的工作区间,越快结束的放在前面(r不相等时,小r放前面,r相等时,大l放前面),按顺序处理结构体p里面的牛,能安排队列q中的牛就优先安排,安排不了就加牛
#include<iostream> #include<string.h> #include<string> #include<algorithm> #include<queue> #include<set> #define ll long long using namespace std; struct node int id; int l; int r; p[1000005],temp,now; int vis[1000005]; bool cmp(node x,node y)//将开始处理时间早,处理时间短的放前面 if(x.l!=y.l) return x.l<y.l; else return x.r<y.r; bool operator < (node x,node y)//二次处理的时候,将结束时间早,开始时间晚的放前面 if(x.r!=y.r) return x.r>y.r; else return x.l<y.l; priority_queue<node>q; int main() int n; while(~scanf("%d",&n)) for(int i=0;i<n;i++) scanf("%d%d",&p[i].l,&p[i].r); p[i].id=i; sort(p,p+n,cmp); int cnt=1; q.push(p[0]); vis[p[0].id]=1; for(int i=1;i<n;i++) if(!q.empty()&&q.top().r<p[i].l) vis[p[i].id]=vis[q.top().id]; q.pop(); else cnt++; vis[p[i].id]=cnt; q.push(p[i]); printf("%d\n",cnt); for(int i=0;i<n;i++) printf("%d\n",vis[i]); while(!q.empty()) q.pop(); return 0;
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