Codeforces Round #298 (Div. 2) D. Handshakes 构造
D. Handshakes
Time Limit: 1 Sec Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/534/problem/D
Description
On February, 30th n students came in the Center for Training Olympiad Programmers (CTOP) of the Berland State University. They came one by one, one after another. Each of them went in, and before sitting down at his desk, greeted with those who were present in the room by shaking hands. Each of the students who came in stayed in CTOP until the end of the day and never left.
At any time any three students could join together and start participating in a team contest, which lasted until the end of the day. The team did not distract from the contest for a minute, so when another student came in and greeted those who were present, he did not shake hands with the members of the contest writing team. Each team consisted of exactly three students, and each student could not become a member of more than one team. Different teams could start writing contest at different times.
Given how many present people shook the hands of each student, get a possible order in which the students could have come to CTOP. If such an order does not exist, then print that this is impossible.
Please note that some students could work independently until the end of the day, without participating in a team contest.
Input
Output
If the sought order of students exists, print in the first line "Possible" and in the second line print the permutation of the students' numbers defining the order in which the students entered the center. Number i that stands to the left of number j in this permutation means that the i-th student came earlier than the j-th student. If there are multiple answers, print any of them.
If the sought order of students doesn't exist, in a single line print "Impossible".
Sample Input
2 1 3 0 1
9
0 2 3 4 1 1 0 2 2
Sample Output
4 5 1 3 2
Possible
7 5 2 1 6 8 3 4 9
HINT
In the first sample from the statement the order of events could be as follows:
- student 4 comes in (a4 = 0), he has no one to greet;
- student 5 comes in (a5 = 1), he shakes hands with student 4;
- student 1 comes in (a1 = 2), he shakes hands with two students (students 4, 5);
- student 3 comes in (a3 = 3), he shakes hands with three students (students 4, 5, 1);
- students 4, 5, 3 form a team and start writing a contest;
- student 2 comes in (a2 = 1), he shakes hands with one student (number 1).
In the second sample from the statement the order of events could be as follows:
- student 7 comes in (a7 = 0), he has nobody to greet;
- student 5 comes in (a5 = 1), he shakes hands with student 7;
- student 2 comes in (a2 = 2), he shakes hands with two students (students 7, 5);
- students 7, 5, 2 form a team and start writing a contest;
- student 1 comes in(a1 = 0), he has no one to greet (everyone is busy with the contest);
- student 6 comes in (a6 = 1), he shakes hands with student 1;
- student 8 comes in (a8 = 2), he shakes hands with two students (students 1, 6);
- student 3 comes in (a3 = 3), he shakes hands with three students (students 1, 6, 8);
- student 4 comes in (a4 = 4), he shakes hands with four students (students 1, 6, 8, 3);
- students 8, 3, 4 form a team and start writing a contest;
- student 9 comes in (a9 = 2), he shakes hands with two students (students 1, 6).
In the third sample from the statement the order of events is restored unambiguously:
- student 1 comes in (a1 = 0), he has no one to greet;
- student 3 comes in (or student 4) (a3 = a4 = 1), he shakes hands with student 1;
- student 2 comes in (a2 = 2), he shakes hands with two students (students 1, 3 (or 4));
- the remaining student 4 (or student 3), must shake one student's hand (a3 = a4 = 1) but it is impossible as there are only two scenarios: either a team formed and he doesn't greet anyone, or he greets all the three present people who work individually.
题意
给你n个人,然后告诉你这n个人进去的时候分别里面坐着有d[i]个人
题解:
直接暴力特判就好了,我们拿一个flag记录这个房间里面有多少个人
代码:
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
//const int inf=0x7fffffff; //无限大
const int inf=0x3f3f3f3f;
/*
inline ll read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int buf[10];
inline void write(int i) {
int p = 0;if(i == 0) p++;
else while(i) {buf[p++] = i % 10;i /= 10;}
for(int j = p-1; j >=0; j--) putchar('0' + buf[j]);
printf("\n");
}
*/
//**************************************************************************************
struct node
{
int x,y;
};
bool cmp(node a,node b)
{
return a.x<b.x;
}
node a[maxn];
map<int,int> H;
vector<int> aa[maxn];
int dp[maxn];
int main()
{
int n;
scanf("%d",&n);
for(int i=;i<n;i++)
{
scanf("%d",&a[i].x);
H[a[i].x]++;
aa[a[i].x].push_back(i+);
}
int now=;
int tot=;
vector<int> ans;
while(tot<n&&now>=)
{
if(H[now]>)
{
H[now]--;
ans.push_back(aa[now][dp[now]]);
dp[now]++;
now++;
}
else
now-=;
}
if(ans.size()!=n)
puts("Impossible");
else
{
puts("Possible");
for(int i=;i<ans.size();i++)
printf("%d ",ans[i]);
}
}
Codeforces Round #298 (Div. 2) D. Handshakes 构造的更多相关文章
- Codeforces Round #298 (Div. 2) D. Handshakes [贪心]
传送门 D. Handshakes time limit per test 1 second memory limit per test 256 megabytes input standard in ...
- Codeforces Round #298 (Div. 2)--D. Handshakes
#include <stdio.h> #include <algorithm> #include <set> using namespace std; #defin ...
- Codeforces Round #298 (Div. 2) A、B、C题
题目链接:Codeforces Round #298 (Div. 2) A. Exam An exam for n students will take place in a long and nar ...
- Codeforces Round #298 (Div. 2) E. Berland Local Positioning System 构造
E. Berland Local Positioning System Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.c ...
- Codeforces Round #298 (Div. 2) A. Exam 构造
A. Exam Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/534/problem/A Des ...
- Codeforces Round #339 (Div. 1) C. Necklace 构造题
C. Necklace 题目连接: http://www.codeforces.com/contest/613/problem/C Description Ivan wants to make a n ...
- CodeForces Round #298 Div.2
A. Exam 果然,并没有3分钟秒掉水题的能力,=_=|| n <= 4的时候特判.n >= 5的时候将奇数和偶数分开输出即可保证相邻的两数不处在相邻的位置. #include < ...
- Codeforces Round #181 (Div. 2) A. Array 构造
A. Array 题目连接: http://www.codeforces.com/contest/300/problem/A Description Vitaly has an array of n ...
- Codeforces Round #306 (Div. 2) ABCDE(构造)
A. Two Substrings 题意:给一个字符串,求是否含有不重叠的子串"AB"和"BA",长度1e5. 题解:看起来很简单,但是一直错,各种考虑不周全, ...
随机推荐
- WGS84转大地2000
1.创建自定义地理(坐标)变换: 2.选择源坐标系和目标坐标系: 3.自定义地理转换方法,选择COORDINATE_FRAME; 4.选择投影工具: 5.在地理变换处选择刚才自定义变换.
- 微信小程序入坑之自定义组件
前言 最近接触微信小程序,再次之前公司用的前端框架是vue ,然后对比发现,开发小程序是各种限制,对于开发者非常不友好.各种槽点太多,完全吐槽不过来,所以在此不多说,打算下次专门写一篇文章吐槽一下.本 ...
- day09作业
一.填空题 1.方法 2.堆内存 3.构造方法 4.this 5.this 6.static 7.使用类名进行访问 8.package import class 9.关键字 10.lang 二.选择题 ...
- supervisor的安装和配置
1. 安装 yum install supervisor 2.配置 [unix_http_server] file=/tmp/supervisor.sock ;UNIX socket 文件,super ...
- maven网址
http://www.yiibai.com/maven/maven_environment_setup.html
- sql 修改列名及表名 sp_rename
因需求变更要改表的列名,平常都是跑到Enterprise manager中选取服务器->数据库->表,然后修改表,这样太麻烦了,查了一下,可以用script搞定, 代码如下: EXEC s ...
- servlet 学习笔记(三)
同一用户的不同页面共享数据有以下四种方法: 1.sendRedirect()跳转 2.session技术 3.隐藏表单提交(form) 4. cookie技术(小甜饼) --------------- ...
- mongo blancer
在 sharded cluster 体系结构中,Balancer 进程的作用是转移数据,当一个 shard 中的数据比其它 shard 的数据多并达到一定条件时,Balancer 进程触发. 为了减少 ...
- SRILM语言模型格式解读
先看一下语言模型的输出格式 \data\ ngram = ngram = ngram = \-grams: -5.24036 'cause -0.2084827 -4.675221 'em -0.22 ...
- linux 101 hacks 7crontab
技巧 74: crontab 书上的这一段我删了,重写一下,用的是ubuntu 16.04系统 参考 每天一个 linux 命令(50):crontab 命令 crond 服务 安装 crontab: ...