F. Contestants Ranking
time limit per test

1 second

memory limit per test

24 megabytes

input

standard input

output

standard output

Ahmad is one of the best students in HIAST, and also a very good problems Solver. In the time you will spend reading this problem statement Ahmad would have solved a problem. Maybe, even two... Ahmad participated so many times in programming contest (ACM-HCPC) with several teams. many of the former and present contestants at HIAST have known Ahmad for quite a few years. Some of them are proud to say that they either played in the same team with him or played in the same team with one of his teammates... Let us define ranking number as follows. Ahmad's ranking is 0, for people who played in the same team with him, the ranking is 1. For people who never played with Ahmad but played in the same team with one or more of his teammates, the ranking is 2, and so on. Your task is to automate the process of calculating the ranking numbers for each contestant at HIAST.

Input

The first line of input file contains the number of test cases (0<T<10). Each test case will begin with one line containing the number of teams (1<N ≤ 100). In each of the following N lines you are given the names of the three members of the corresponding team. Each name is a nonempty string contains only English letters starts with capital letter, and its length is at most 20 symbols. Same student can be found in more than one team, but each student should have only one rank. Ahmad will be found in one team at least. The first letter of a name is capital and the other letters are lowercase and each name will consist of only one word.

Output

For each test case output a line with the number of contestants, then for each contestant output a line with his name and his ranking. If the ranking is undefined, output “undefined” instead of it. The contestants must be ordered by rank from 0 to undefined and then lexicographical by name.

Examples
Input
2
1
Ahmad Mousaab Khalid
7
Ahmad Mousaab Khalid
Ali Mousaab Nizar
Ali Bassel Nizar
Kassem Ahmad Mousaab
Saeed Kassem Fadel
Salwa Saeed Samer
Mona Abdo Qussi
Output
3
Ahmad 0
Khalid 1
Mousaab 1
14
Ahmad 0
Kassem 1
Khalid 1
Mousaab 1
Ali 2
Fadel 2
Nizar 2
Saeed 2
Bassel 3
Salwa 3
Samer 3
Abdo undefined
Mona undefined
Qussi undefined
每次先确定一个等级在查找下个等级,没有等级的视为undefined
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <map>
#include <set>
#include <vector>
#include <algorithm>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
string s[][],name="Ahmad";
map<string,int>m;
set<string>ans[],pos;
int t,n;
void check(int x,int y,int z)
{
if(ans[x].count(s[y][z]))
{
for(int i=;i<=;i++)
{
       if(m[s[y][(z+i)%]]==)
{
ans[x+].insert(s[y][(z+i)%]);
m[s[y][(z+i)%]]=;
}
}
}
}
void check_undefined(int n)
{
for(int i=;i<n;i++)
{
for(int j=;j<;j++)
{
if(m[s[i][j]]==) pos.insert(s[i][j]),m[s[i][j]]=;
}
}
}
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=;i<n;i++)
{
for(int j=;j<;j++)
{
cin>>s[i][j];
}
}
m[name]=;
ans[].insert(name);
for(int i=;i<n;i++)
{
for(int j=;j<n;j++)
{
for(int k=;k<;k++)
{
check(i,j,k);
}
}
}
check_undefined(n);
printf("%d\n",m.size());
for(int i=;i<=n;i++)
{
if(ans[i].size()==) break;
for(set<string>::iterator it=ans[i].begin();it!=ans[i].end();it++)
cout<<*it<<' '<<i<<endl;
}
for(set<string>::iterator it=pos.begin();it!=pos.end();it++)
cout<<*it<<" undefined"<<endl;
for(int i=;i<;i++)
ans[i].clear();
pos.clear();
m.clear();
}
return ;
}

Gym 100952 F. Contestants Ranking的更多相关文章

  1. Gym 100952F&&2015 HIAST Collegiate Programming Contest F. Contestants Ranking【BFS+STL乱搞(map+vector)+优先队列】

    F. Contestants Ranking time limit per test:1 second memory limit per test:24 megabytes input:standar ...

  2. codeforces gym 100952 A B C D E F G H I J

    gym 100952 A #include <iostream> #include<cstdio> #include<cmath> #include<cstr ...

  3. Gym 100952 H. Special Palindrome

    http://codeforces.com/gym/100952/problem/H H. Special Palindrome time limit per test 1 second memory ...

  4. Gym 100952 D. Time to go back(杨辉三角形)

    D - Time to go back Gym - 100952D http://codeforces.com/gym/100952/problem/D D. Time to go back time ...

  5. Gym 100952 G. The jar of divisors

    http://codeforces.com/gym/100952/problem/G G. The jar of divisors time limit per test 2 seconds memo ...

  6. Gym 100952 D. Time to go back

    http://codeforces.com/gym/100952/problem/D D. Time to go back time limit per test 1 second memory li ...

  7. Gym 100952 C. Palindrome Again !!

    http://codeforces.com/gym/100952/problem/C C. Palindrome Again !! time limit per test 1 second memor ...

  8. Gym 100637F F. The Pool for Lucky Ones

    F. The Pool for Lucky Ones Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/10 ...

  9. codeforces Gym 100187F F - Doomsday 区间覆盖贪心

    F. Doomsday Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100187/problem/F ...

随机推荐

  1. centos7 初始化安装

    CENTOS7 初装 一.分区 挂载路径 格式 容量 / xfs 102400 swap 等同内存大小 /home xfs 剩余 二.时区 Asia/Shanghai 三.安装包选择 选择最小化安装 ...

  2. NodeJS学习笔记 (32)安全加密-tls

    https://github.com/chyingp/nodejs-learning-guide

  3. NodeJS学习笔记 (27)实用工具模块-util(ok)

    debuglog(section) 很有用的调试方法.可以通过 util.debuglog(name) 来创建一个调试fn,这个fn的特点是,只有在运行程序时候,声明环境变量NODE_DEBUG=na ...

  4. win7 ssd评分降为5.9的诡异问题解决方法

    某一天偶然发现win7的系统评分里面,磁盘一项由之前的7.9降到5.9了,SSD早听说会有降速的问题,但无论如何降,也不至于被降到5.9分这一机械硬盘普遍的分数. 百度搜了,Google搜了,中文搜了 ...

  5. ArcGIS Engine获得要素的中心点坐标

    IPoint centerPoint =new PointClass();//获得要素的中心点 IArea pArea = pFeature.Shape as IArea; pArea.QueryCe ...

  6. iOS 画圆形头像

    demo下载地址:http://pan.baidu.com/s/1mgBf6YG _logoImageView.image = [self getEllipseImageWithImage:[UIIm ...

  7. iOS学习9_事件分发&amp;响应链

    iOS的三种事件:触摸事件/运动事件/远程控制事件 typedef enum { UIEventTypeTouches, UIEventTypeMotion, UIEventTypeRemoteCon ...

  8. vue --- axios发post请求后台接收不到参数的三种解决方案

    最近用vue  做项目使用axios 发送post 请求时遇到了前端传数据后端接收不到的情况: 后来仔细对比发现axios传值是这样的: 而 ajax 传值是这样的: 一个 Request Paylo ...

  9. maven pom下载不了

    Downloading: http://repo.maven.apache.org/maven2/org/apache/maven/plugins/maven-clean-plugin/2.4.1/m ...

  10. 【搭建Saltstack运维工具】

    目录 所谓Salt 开始搭建 配置接受密钥 salt命令 YAML详解 目标定位字符串 state模块定义主机状态 Salt采集静态信息之GrainsSalt @(Saltstack) *** 所谓S ...