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
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
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
Lines 2..N+1: Line i+1 describes cow i's milking interval with two space-separated integers.
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
Lines 2..N+1: Line i+1 describes the stall to which cow i will be assigned for her milking period.
5 1 10 2 4 3 6 5 8 4 7Sample Output
4 1 2 3 2 4Hint
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:给出每个奶牛的挤奶时间,一台机器只能挤一头奶牛,至少要多少太机器,输出奶牛使用机器的编号。思路:按照每一头奶牛的开始时间,用优先队列,将奶牛入队,如果前者开始时间大于后者,则需要添加机器。注释理解
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
int flag[50005];
struct node
{
int cow;
int tim;
int num;//记录第几个奶牛
friend bool operator <(node a,node b)
{
if(a.tim==b.tim)//时间相同时,结束时间从小到大排,保证优先只用一个机器
{
return a.cow<b.cow;
}
return a.tim>b.tim;
}
}m[50005];
bool cmp(node x,node y)//排序
{
if(x.cow!=y.cow)
return x.cow<y.cow;
else
return x.tim<y.tim;
}
priority_queue<node>temp;
int main()
{
int n,i,sum;
while(cin>>n)
{
for(i=0;i<n;i++)
{
cin>>m[i].cow>>m[i].tim;
m[i].num=i;
}
sort(m,m+n,cmp);
sum=1;
temp.push(m[0]);
flag[m[0].num]=1;
for(i=1;i<n;i++)
{
if(!temp.empty()&&temp.top().tim<m[i].cow)
{
flag[m[i].num]=flag[temp.top().num];
temp.pop();
}
else
{
sum+=1;//第几个机器
flag[m[i].num]=sum;
}
temp.push(m[i]);
}
cout<<sum<<endl;
for(i=0;i<n;i++)
{
cout<<flag[i]<<endl;
}
while(!temp.empty())
{
temp.pop();
}
}
return 0;
}