B. F1 Champions
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Formula One championship consists of series of races called Grand Prix. After every race drivers receive points according to their final position. Only the top 10 drivers receive points in the following order 25, 18, 15, 12, 10, 8, 6, 4, 2, 1. At the conclusion
of the championship the driver with most points is the champion. If there is a tie, champion is the one with most wins (i.e. first places). If a tie still exists, it is chosen the one with most second places, and so on, until there are no more place to use
for compare.

Last year another scoring system was proposed but rejected. In it the champion is the one with most wins. If there is tie, champion is the one with most points. If a tie still exists it is proceeded the same way as in the original scoring system, that is comparing
number of second, third, forth, and so on, places.

You are given the result of all races during the season and you are to determine the champion according to both scoring systems. It is guaranteed, that both systems will produce unique champion.

Input

The first line contain integer t (1 ≤ t ≤ 20),
where t is the number of races. After that all races are described one by one. Every race description start with an integer n (1 ≤ n ≤ 50)
on a line of itself, where n is the number of clasified drivers in the given race. After thatn lines
follow with the classification for the race, each containing the name of a driver. The names of drivers are given in order from the first to the last place. The name of the driver consists of lowercase and uppercase English letters and has length at most 50
characters. Comparing of names should be case-sensetive.

Output

Your output should contain exactly two line. On the first line is the name of the champion according to the original rule, and on the second line the name of the champion according to the alternative rule.

Examples
input
3
3
Hamilton
Vettel
Webber
2
Webber
Vettel
2
Hamilton
Vettel
output
Vettel
Hamilton
input
2
7
Prost
Surtees
Nakajima
Schumacher
Button
DeLaRosa
Buemi
8
Alonso
Prost
NinoFarina
JimClark
DeLaRosa
Nakajima
Patrese
Surtees
output
Prost
Prost

结构体排序
#include <iostream>
#include <algorithm>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <stdio.h>
#include <map>
#include <string> using namespace std;
int n,m;
struct Node
{
int score;
int r[55];
string name;
}a[1005];
int cmp1(Node a,Node b)
{
if(a.score==b.score)
{
for(int i=1;i<50;i++)
{
if(a.r[i]==b.r[i])
continue;
return a.r[i]>b.r[i];
}
}
return a.score>b.score;
}
int cmp2(Node a,Node b)
{
if(a.r[1]==b.r[1])
{
if(a.score==b.score)
{
for(int i=2;i<50;i++)
{
if(a.r[i]==b.r[i])
continue;
return a.r[i]>b.r[i];
}
}
return a.score>b.score;
}
return a.r[1]>b.r[1];
}
map<string,int> mm; int num[55]={0,25,18,15,12,10,8,6,4,2,1};
int main()
{
scanf("%d",&n);
string s;
int cnt=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&m);
for(int j=1;j<=m;j++)
{
cin>>s;
if(!mm.count(s))
mm[s]=cnt++;
a[mm[s]].score+=num[j];
a[mm[s]].r[j]++;
a[mm[s]].name=s;
}
}
sort(a,a+cnt,cmp1);
cout<<a[0].name<<endl;
sort(a,a+cnt,cmp2);
cout<<a[0].name<<endl;
return 0;
}

CodeForces 24B F1 Champions(排序)的更多相关文章

  1. 24B F1 Champions

    传送门 题目 Formula One championship consists of series of races called Grand Prix. After every race driv ...

  2. Codeforces 510C (拓扑排序)

    原题:http://codeforces.com/problemset/problem/510/C C. Fox And Names time limit per test:2 seconds mem ...

  3. E - E CodeForces - 1100E(拓扑排序 + 二分)

    E - E CodeForces - 1100E 一个n个节点的有向图,节点标号从1到n,存在m条单向边.每条单向边有一个权值,代表翻转其方向所需的代价.求使图变成无环图,其中翻转的最大边权值最小的方 ...

  4. CodeForces 721C Journey(拓扑排序+DP)

    <题目链接> 题目大意:一个DAG图有n个点,m条边,走过每条边都会花费一定的时间,问你在不超过T时间的条件下,从1到n点最多能够经过几个节点. 解题分析:对这个有向图,我们进行拓扑排序, ...

  5. Divide by three, multiply by two CodeForces - 977D (思维排序)

    Polycarp likes to play with numbers. He takes some integer number xx, writes it down on the board, a ...

  6. Selling Souvenirs CodeForces - 808E (分类排序后DP+贪心)

    E. Selling Souvenirs time limit per test 2 seconds memory limit per test 256 megabytes input standar ...

  7. Cinema CodeForces - 670C (离散+排序)

    Moscow is hosting a major international conference, which is attended by n scientists from different ...

  8. CodeForces - 589B(暴力+排序)

    Dasha decided to bake a big and tasty layer cake. In order to do that she went shopping and bought n ...

  9. CodeForces - 919D Substring (拓扑排序+dp)

    题意:将一个字符串上的n个字符视作点,给出m条有向边,求图中路径上最长出现的相同字母数. 分析:首先如果这张图中有环,则可以取无限大的字符数,在求拓扑排序的同时可以确定是否存在环. 之后在拓扑排序的结 ...

随机推荐

  1. svn add xxx.txt 提示A (bin) xxx.txt

    [root@NGINX-APACHE-SVN iptables]# svn ci -m "add iptables.txt" Adding (bin) iptables/iptab ...

  2. Linux之系统管理员笔记

    1.查看最近一次启动时间 who -b system boot -- : 2.who命令实现带有“表头”的查询结果 who -H NAME LINE TIME COMMENT root pts/ -- ...

  3. pc或者微信上用pdf.js在线预览pdf和word

    最近项目要求pdf和word可以在线预览功能,pc端还好解决,但是微信端就有点坑了,pc端原来的思路是将文件转成base64,然后用html格式显示 ,但是微信端不支持, 这种方式就pass掉了,谷歌 ...

  4. js eval深入

    在JS中将JSON的字符串解析成JSON数据格式,一般有两种方式: 1.一种为使用eval()函数. 2. 使用Function对象来进行返回解析. 使用eval函数来解析,并且使用jquery的ea ...

  5. Makefile 12——改善编译效率

    从Makefile的角度看,一个可以改善编译效率的地方与其中的隐式规则有关.为了了解make的隐式规则,我们把之前的simple项目的Makefile做一点更改,删除生成.o文件的规则(与隐式规则相对 ...

  6. 基于jquery的适合电子商务网站首页的图片滑块

    今天给大家分享一款基于Sequence.js 的图片滑动效果,特别适合电子商务网站或者企业产品展示功能.带有图片缩率图,能够呈现全屏图片浏览效果.结合 CSS3 Transition 实现响应式的滑块 ...

  7. oracle获取SID

    windows 下查看注册表 开始 输入regedit 查看HKEY_LOCAL_MACHINE\SOFTWARE\ORACLE\KEY_OraDb11g_home1\ORACLE_SID就是 lin ...

  8. DRBD(Distributed Replicated Block Device) 分布式块设备复制 进行集群高可用方案

    DRBD是一个用软件实现的.无共享的.服务器之间镜像块设备内容的存储复制解决方案. 外文名 DRBD drbdadm 高级管理工具 drbdsetup 置装载进kernel的DRBD模块 drbdme ...

  9. word 操作教程

    http://blog.163.com/haolongqin@126/blog/static/10999842220159993540527/ https://blog.csdn.net/ibigpi ...

  10. 【BZOJ】1024: [SCOI2009]生日快乐(dfs)

    http://www.lydsy.com/JudgeOnline/problem.php?id=1024 果然现在弱到连搜索都不会了么..... 一直想二分...但是无论如何也推不出怎么划分... Q ...