poj 3190 Stall Reservations (贪心+优先队列）

链接：http://poj.org/problem?id=3190

Stall Reservations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6645		Accepted: 2401		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

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:

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.

题意：N只牛，用挤奶器挤奶，时间为s-e，试计算最少用多少个挤奶器能挤完这n只牛？

思路：贪心，先按开始时间从小到大排序，这样能够保证不会有奶牛的挤奶时间在上一只前面，然后建立优先队列，按结束时间从小到大排序，然后依次向队列中加入n只奶牛，同时拿出一只奶牛，如果拿出的奶牛结束时间小于当前加入奶牛的开始时间，表示这两头奶牛可以共用一个挤奶器，反之不能

贪心思路比较容易想到，但是优先队列没有想到，看到别人用优先队列就恍然大悟了

代码：

#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<queue>
#include<algorithm>
using namespace std;
const int maxn = 50000 + 10;

struct node {
int s;
int e;
int k;
};

int n;
node cow[maxn];
int vis[maxn];//挤奶器

bool cmp1(node a, node b) {
if (a.s == b.s)
return a.e < b.e;
else return a.s < b.s;
}
bool operator < (const node &a, const node &b)//产奶结束越早的越优先
{
if (a.e == b.e)
return a.s>b.s;
return a.e>b.e;
}
int main() {
while (scanf("%d", &n) != EOF) {
memset(vis, 0, sizeof(vis));
for (int i = 1; i <= n;i++) {
scanf("%d %d", &cow[i].s, &cow[i].e);
cow[i].k = i;
}
sort(cow + 1, cow + n + 1, cmp1);
priority_queue<node>Q;//按结束时间排序
Q.push(cow[1]);
vis[cow[1].k] = 1;
int j = 1;
for (int i = 2; i <= n;i++) {
node neww = Q.top();
if (cow[i].s > neww.e) {//可以同用一个挤奶器
vis[cow[i].k] = vis[neww.k];
Q.pop();
}
else {//不能同用一个挤奶器
j++;
vis[cow[i].k] = j;
}
Q.push(cow[i]);
}
cout << j << endl;
for (int i = 1; i <= n; i++)cout << vis[i] << endl;
}

return 0;
}