poj3190(Stall Reservations)

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.

将n个区间分成最少的组，使每个组的区间没有重合。按区间左端点从小到大排序，用优先队列记录stall及其最右端端点，优先队列按照右端点从小到大排序。每次取出右端点最小的stall，如果当前区间与之有重合，则新增一个stall，否则更新该stall右端点。

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
int dx[] = {0, 1, 0, -1}, dy[] = {-1, 0, 1, 0};
const int maxn = 50010;

struct stall
{
    int name, last;
    bool operator < (const stall& b) const
    {
        return last > b.last;
    }
};

struct cow
{
    int name, start, over;
    bool operator < (const cow& b) const
    {
        return start < b.start;
    }
};

int main()
{
    int n;
    cin >> n;
    cow s[maxn];
    for (int i = 0; i < n; ++i)
    {
        scanf("%d%d", &s[i].start, &s[i].over);
        s[i].name = i;
    }
    sort(s, s+n);
    priority_queue<stall> q;
    int ans = 1, num[maxn];
    num[s[0].name] = 1;
    q.push({1, s[0].over});
    for (int i = 1; i < n; ++i)
    {
        stall tmp = q.top();
        q.pop();
        if (s[i].start > tmp.last)
        {
            num[s[i].name] = tmp.name;
            tmp.last = s[i].over;
            q.push(tmp);
        }
        else
        {
            num[s[i].name] = ++ans;
            q.push(tmp);
            q.push({ans, s[i].over});
        }
    }
    printf("%d\n", ans);
    for (int i = 0; i < n; ++i)
        printf("%d\n", num[i]);
    return 0;
}