zoj3317Murder in Restaurant

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本文介绍了一种通过模拟和排序解决餐厅凶案中房间分配问题的方法。输入包含租客数量及租住天数，输出每个租客所住房间号。采用结构体存储租客信息并实现两次排序。

链接1：

http://openoj.awaysoft.com:8080/judge/contest/view.action?cid=269#problem/C

链接2：

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3317

Murder in Restaurant

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%lld & %lluSubmit Status Practice ZOJ 3317

Description

Murder in closet happened again in a small restaurant and Conan went there to collect evidence with Kogoro. After they reached the restaurant, they got a list of renters that lived in this restaurant recently. Yet the list was so long that they couldn't get the useful information from it. As an assistant of Conan, you must have been ready to help him to get the information on which rooms the renters have lived in.

Input

There are no more than 20 test cases. The first line of each test case contains two integers n and m, indicating the number of renters and the rooms of the restaurant (0 < n, m <= 100). The i-th line of the next n lines contains two integers t1 and t2, the day when they wanted to check in and to leave (0 < t1 < t2 <= 1000).

Each renter rends exactly one room and their check-in days are distinct. Each time a renter came, the owner would give him/her an available room with minimum room number if there were still empty rooms. Otherwise the renter would leave at once and never come back again. Note that rooms are always returned in morning and rented in afternoon. The input is ended with two zeroes.

Output

For each test case, output n lines of integers. The i-th integer indicates the room number of the i-th renter. The rooms are numbered from 1. If someone didn't live in this restaurant, output 0 in the corresponding line.

Sample Input

2 5

1 3

2 4

4 2

1 5

2 3

3 5

4 5

0 0

Sample Output

思路：模拟数组和排序：

Memory：188KB

Time：0ms

Length:749B


#include<stdio.h>

#include<string.h>

#include<algorithm>

#include<iostream>

using namespace std;

struct Renter//客人信息 

{

int s;

int t;

int id; 

int rn;

}r[105];

int cmp1(Renter a, Renter b )

{

    return a.s<b.s;

}

int cmp2(Renter a,Renter b)

{

return  a.id<b.id; 

}


int main()

{

    int n,m,i,j;

    int room[105];

    while(~scanf("%d%d",&n,&m)&&!(n==0&&m==0))

    {memset(room,0,sizeof(room));


    for(i=1;i<=n;i++)

    {scanf("%d%d",&r[i].s,&r[i].t);

     r[i].rn=0;//将客人入住情初始化为零 

     r[i].id=i;

    }

    sort(r+1,r+1+n,cmp1);

    for(i=1;i<=n;i++)//n个顾客在m个房建中一一做选择 

    for(j=1;j<=m;j++)

    {

    if(room[j]<=r[i].s)

    {

    r[i].rn=j;

    room[j]=r[i].t;

    break;

    }

    }

    sort(r+1,r+1+n,cmp2);//记住这里要按照原来的次序输出 

    for(i=1;i<=n;i++)

    printf("%d\n",r[i].rn);

    }

    return 0;

}