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如题：http://poj.org/problem?id=3190

Stall Reservations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3380		Accepted: 1207		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 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

USACO 2006 February Silver

题目很容易就能懂，给出几个区间，不能重叠，问需要的最小stall。

首先对输入数据排序，按左端点小->大，然后开始遍历，每次插入一个后，用优先队列维护。维护的是右端点最小。

不难发现，每次插入最左端，那么只用判断右端点，每次选择右端点最小的那个插入，最终方案最优，如果最小的那个都有重叠区间，那么stall的数量只能再+1了。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<queue>
using namespace std;
#define min(a,b)(a< b?a:b)
struct node
{
int l,r;
int pos;
};
node Line[50005];

int stall[50005];
int cmp(node &a,node &b)
{
if(a.l!=b.l)
return a.l<b.l;
return a.r<b.r;
}

class cmp1
{
public:
bool operator()(node&a,node&b)
{
  return a.r>b.r;
}
};
int main()
{
// freopen("C:\\1.txt","r",stdin);
int N,num=1;
scanf("%d",&N);
int i;
for(i=0;i<N;i++)
{
  scanf("%d%d",&Line[i].l,&Line[i].r);
  Line[i].pos=i+1;
}
sort(Line,Line+N,cmp);
priority_queue<node,vector<node>,cmp1>que;
que.push(Line[0]);
stall[Line[0].pos]=1;
i=1;
while(i<N)
{
  node p=que.top();
  if(Line[i].l>p.r)
  {
   que.pop();
   stall[Line[i].pos]=stall[p.pos];
  }
  else if(Line[i].l<=p.r)
  {
   num++;
   stall[Line[i].pos]=num;
  }
  que.push(Line[i]);
  i++;
}
printf("%d\n",num);
for(i=1;i<=N;i++)
  printf("%d\n",stall[i]);
}