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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:
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. Source |
额，贪心题=。=，STL里的优先队列
贪心策略：对奶牛挤奶开始时间由大到小排序，这样保证不会考虑有奶牛的开始挤奶时间在前面；
对入队的元素按挤奶结束时间时间由小到大排序，这样保证第一次就可以判定用不用添加槽=。=
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
#include<queue>
using namespace std;
const int maxn=1000000+10;
struct node{
int st,ed;
int pos;//pos记录是第几个奶牛
bool operator<(const node &a)const
{
if(ed==a.ed)
return st>a.st;
return ed>a.ed;
}
}a[maxn];
int used[maxn];//used记录奶牛挤奶的位置=。=
int cmp(node l1,node l2)
{
if(l1.st==l2.st)
return l1.ed<l2.ed;
return l1.st<l2.st;
}
int main()
{
int n;
while(~scanf("%d",&n))
{
priority_queue<node>q;
for(int i=1;i<=n;i++)
{
scanf("%d%d",&a[i].st,&a[i].ed);
a[i].pos=i;
}
sort(a+1,a+n+1,cmp);
q.push(a[1]);
int ans=1;
used[a[1].pos]=1;
for(int i=2;i<=n;i++)
{
if(!q.empty()&&q.top().ed<a[i].st)//判定是否符合条件
{
used[a[i].pos]=used[q.top().pos];
q.pop();
}
else//不符合条件槽的数量加一，同时这个奶牛应该在新加的槽
{
ans++;
used[a[i].pos]=ans;
}
q.push(a[i]);
}
printf("%d\n",ans);
for(int i=1;i<=n;i++)
printf("%d\n",used[i]);
}
return 0;
}
POJ 3190 Stall Reservations(贪心+优先队列优化）,布布扣,bubuko.com
POJ 3190 Stall Reservations(贪心+优先队列优化）
原文：http://blog.csdn.net/u013582254/article/details/38311935