Time Limit: 1000MS
Memory Limit: 65536K
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:
Other outputs using the same number of stalls are possible.
Source
USACO 2006 February Silver
解题分析
所有奶牛都必须挤奶。到了一个奶牛的挤奶开始时间,就必须为这个奶牛找畜栏。因此按照奶牛的开始时间逐个处理它们,是必然的。
1)把所有奶牛按开始时间从小到大排序。
2)为第一头奶牛分配一个畜栏。
3)依次处理后面每头奶牛
- 若
E(x)<S(i) ,则不用分配新畜栏,i 可进入x ,并修改E(x) 为E(i) 。 - 若
E(x)>=S(i) ,则分配新畜栏y ,记E(y)=E(i) 。
直到所有奶牛处理结束。
需要用优先队列存放已经分配的畜栏,并使得结束时间最早的畜栏始终位于队列头部。
证明:
由于按开始时间的顺序处理奶牛是必然,且按该算法,为奶牛
复杂度:
AC代码
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
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 et;
int No;
bool operator < (const Stall &s)
const{return et > s.et;}
Stall(int e,int n):et(e),No(n){}
};
priority_queue<Stall>pq;
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;
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.et<cows[i].a){
pq.pop();
pq.push(Stall(cows[i].b,st.No));
pos[cows[i].No]=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;
}