Stall Reservations POJ-3190 解题报告_综合

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:

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

Sample Output

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.

Source

USACO 2006 February Silver

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;
}