题目:POJ-2965
http://poj.org/problem?id=3190
Stall Reservations
Time Limit: 1000MS | Memory Limit: 65536K | |||
Total Submissions: 9673 | Accepted: 3383 | 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 10 Stall 1 c1>>>>>>>>>>>>>>>>>>>>>>>>>>> Stall 2 .. c2>>>>>> c4>>>>>>>>> .. .. Stall 3 .. .. c3>>>>>>>>> .. .. .. .. Stall 4 .. .. .. c5>>>>>>>>> .. .. ..Other outputs using the same number of stalls are possible.
Source
解题思路:
贪心算法!!
题目是跟着POJ很好很有层次感做的,在做POJ-1328之前学习了贪心算法,学习过程中做了这道题
基本思路:模拟牛挤奶的顺序,按照结束时间将牛排好序,利用优先队列对栅栏进行判断,随时更新栅栏的结束使用时间,并记录每头牛相应的栅栏号
#include <iostream> #include <cstdio> #include <queue> #include <vector> #include <cstring> #include <string> #include <algorithm> using namespace std; struct Cow { int a,b; int No; bool operator < (const Cow & c) const { return a<c.a; } } cows [50100]; int pos [50100]; struct Stall { int end ; int No; bool operator < (const Stall & s) const { return end>s.end; } Stall (int e ,int n):end(e),No(n){ } }; int main () { int n; scanf ("%d",&n); for (int i=0;i<n;++i) { scanf ("%d%d",&cows[i].a,&cows[i].b); cows[i].No=i; } sort(cows,cows+n); int total=0; priority_queue<Stall> pq; for (int i=0;i<n;++i) { if (pq.empty()) { ++total; pq.push(Stall(cows[i].b,total)); pos[cows[i].No]=total; } else { Stall st = pq.top(); if(st.end<cows[i].a) { pq.pop(); pos[cows[i].No]=st.No; pq.push(Stall(cows[i].b,st.No)); } else { ++total; pq.push(Stall(cows[i].b,total)); pos[cows[i].No]=total; } } } printf ("%d\n",total); for (int i=0;i<n;++i) printf("%d\n",pos[i]); return 0; }