码迷,mamicode.com
首页 > 其他好文 > 详细

Stall Reservations (poj 3190 贪心)

时间:2015-03-04 14:47:44      阅读:100      评论:0      收藏:0      [点我收藏+]

标签:stall reservations   poj 3190   贪心   

Language:
Stall Reservations
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 3394   Accepted: 1215   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: 

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


题意:给出每个奶牛挤奶的时间段,一个机器一次只能对一头奶牛工作,问至少需要多少台机器,并输出每头奶牛使用的机器编号。

思路:先按照每头牛的开始时间从小到大排序,维护一个优先队列,n头牛依次入队,时间结束早的先出队列,比较最早出来的牛的结束时间和当前准备入队的牛的开始时间,若前者小于后者,则当前的牛可以用前面空出来的机器,否者要添加一台新机器。

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 50010
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

struct Cow
{
    int s,t;
    int local;
}cow[maxn];

int cmp(Cow a,Cow b)
{
    if (a.s==b.s) return a.t<b.t;
    return a.s<b.s;
}

priority_queue<Cow>Q;
int ans[maxn];
int n;

bool operator<(Cow a,Cow b)
{
    return a.t>b.t;
}

int main()
{
    int i,j;
    Cow st;
    while (~sf(n))
    {
        FRL(i,0,n)
        {
            sff(cow[i].s,cow[i].t);
            cow[i].local=i;
        }
        sort(cow,cow+n,cmp);
        Q.push(cow[0]);
        int cnt=1;
        ans[ cow[0].local ]=cnt;
        FRL(i,1,n)
        {
            if (!Q.empty()&&Q.top().t<cow[i].s)
            {
                st=Q.top();
                Q.pop();
                ans[ cow[i].local ]=ans[ st.local ];
            }
            else
                ans[ cow[i].local ]=++cnt;
            Q.push(cow[i]);
        }
        printf("%d\n",cnt);
        FRL(i,0,n)
        pf("%d\n",ans[i]);
    }

    return 0;
}
/*
5
1 10
2 4
3 6
5 8
4 7
*/



Stall Reservations (poj 3190 贪心)

标签:stall reservations   poj 3190   贪心   

原文地址:http://blog.csdn.net/u014422052/article/details/44058525

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!