Problem A: Football (aka Soccer) 

The Problem

Football the most popular sport in the world (americans insist to call it "Soccer", but we will call it "Football"). As everyone knows, Brasil is the country that have most World Cup titles (four of them: 1958, 1962, 1970 and 1994). As our national tournament have many teams (and even regional tournaments have many teams also) it's a very hard task to keep track of standings with so many teams and games played!

So, your task is quite simple: write a program that receives the tournament name, team names and games played and outputs the tournament standings so far.

A team wins a game if it scores more goals than its oponent. Obviously, a team loses a game if it scores less goals. When both teams score the same number of goals, we call it a tie. A team earns 3 points for each win, 1 point for each tie and 0 point for each loss.

Teams are ranked according to these rules (in this order):

  1. Most points earned.
  2. Most wins.
  3. Most goal difference (i.e. goals scored - goals against)
  4. Most goals scored.
  5. Less games played.
  6. Lexicographic order.

The Input

The first line of input will be an integer N in a line alone (0 < N < 1000). Then, will follow N tournament descriptions. Each one begins with the tournament name, on a single line. Tournament names can have any letter, digits, spaces etc. Tournament names will have length of at most 100. Then, in the next line, there will be a number T (1 < T <= 30), which stands for the number of teams participating on this tournament. Then will follow T lines, each one containing one team name. Team names may have any character that have ASCII code greater than or equal to 32 (space), except for '#' and '@' characters, which will never appear in team names. No team name will have more than 30 characters.

Following to team names, there will be a non-negative integer G on a single line which stands for the number of games already played on this tournament. G will be no greater than 1000. Then, G lines will follow with the results of games played. They will follow this format:

team_name_1#goals1@goals2#team_name_2

For instance, the following line:

Team A#3@1#Team B

Means that in a game between Team A and Team B, Team A scored 3 goals and Team B scored 1. All goals will be non-negative integers less than 20. You may assume that there will not be inexistent team names (i.e. all team names that appear on game results will have apperead on the team names section) and that no team will play against itself.

The Output

For each tournament, you must output the tournament name in a single line. In the next T lines you must output the standings, according to the rules above. Notice that should the tie-breaker be the lexographic order, it must be done case insenstive. The output format for each line is shown bellow:

[a]) Team_name [b]p, [c]g ([d]-[e]-[f]), [g]gd ([h]-[i])

Where:

  • [a] = team rank
  • [b] = total points earned
  • [c] = games played
  • [d] = wins
  • [e] = ties
  • [f] = losses
  • [g] = goal difference
  • [h] = goals scored
  • [i] = goals against

There must be a single blank space between fields and a single blank line between output sets. See the sample output for examples.

Sample Input

2
World Cup 1998 - Group A
4
Brazil
Norway
Morocco
Scotland
6
Brazil#2@1#Scotland
Norway#2@2#Morocco
Scotland#1@1#Norway
Brazil#3@0#Morocco
Morocco#3@0#Scotland
Brazil#1@2#Norway
Some strange tournament
5
Team A
Team B
Team C
Team D
Team E
5
Team A#1@1#Team B
Team A#2@2#Team C
Team A#0@0#Team D
Team E#2@1#Team C
Team E#1@2#Team D

Sample Output

World Cup 1998 - Group A
1) Brazil 6p, 3g (2-0-1), 3gd (6-3)
2) Norway 5p, 3g (1-2-0), 1gd (5-4)
3) Morocco 4p, 3g (1-1-1), 0gd (5-5)
4) Scotland 1p, 3g (0-1-2), -4gd (2-6) Some strange tournament
1) Team D 4p, 2g (1-1-0), 1gd (2-1)
2) Team E 3p, 2g (1-0-1), 0gd (3-3)
3) Team A 3p, 3g (0-3-0), 0gd (3-3)
4) Team B 1p, 1g (0-1-0), 0gd (1-1)
5) Team C 1p, 2g (0-1-1), -1gd (3-4)
题意: 足球比赛。。给定球队。和比赛的情况。比赛为 队伍1 #队伍1赢球数 @ 队伍2赢球数 # 队伍2 这样的格式。
赢一场得3分 平一场得1分 输了不得分。
最后要输出整个比赛的情况。。(排名 队名 得分 总场次 赢场 平场 输场 净胜球 胜球数 输球数)
注意的一点、整个比赛的情况要按 得分 -》 赢场 -》 净胜球 -》 胜球数 -》总场次少的 -》 队名字典序(不分大小写) 排序输出。。。
思路: 就是字符串处理稍微麻烦。。。其实挺水的 
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
char cn[105];
int tnum;
int cnum;
int tt;
struct Team
{
char name[35];
int f;
int z;
int y;
int s;
int p;
int zq;
int yq;
int sq;
} t[35]; int find(char *team)
{
for (int i = 0; i < tnum; i ++)
{
if(strcmp(t[i].name, team) == 0)
{
return i;
}
}
return -1;
} int cmp(Team a, Team b)
{
if (a.f == b.f)
{
if (a.y == b.y)
{
if (a.zq == b.zq)
{
if (a.yq == b.yq)
{
if (a.z == b.z)
{
if (strcasecmp(a.name, b.name) < 0)
return 1;
if (strcasecmp(a.name, b.name) > 0)
return 0;
}
else
return a.z < b.z;
}
else
return a.yq > b.yq;
}
else
return a.zq > b.zq;
}
else
return a.y > b.y;
}
else
return a.f > b.f;
} int main()
{
scanf("%d", &tt);
getchar();
while (tt --)
{
gets(cn);
memset(t, 0, sizeof(t));
scanf("%d", &tnum);
getchar();
for (int i = 0; i < tnum; i ++)
gets(t[i].name);
scanf("%d", &cnum);
getchar();
while (cnum --)
{
char t1[35];
int t1num = 0;
int t1y = 0;
char t2[35];
int t2num = 0;
int t2y = 0;
char sb;
while ((sb = getchar()) != EOF)
{
if (sb == '#')
break;
t1[t1num ++] = sb;
}
t1[t1num] = '\0';
while ((sb = getchar()) != EOF)
{
if (sb == '@')
break;
t1y = t1y * 10 + sb - '0';
}
while ((sb = getchar()) != EOF)
{
if (sb == '#')
break;
t2y = t2y * 10 + sb - '0';
}
while ((sb = getchar()) != EOF)
{
if (sb == '\n')
break;
t2[t2num ++] = sb;
}
t2[t2num] = '\0';
int t11 = find(t1);
int t22 = find(t2);
t[t11].yq += t1y;
t[t11].sq += t2y;
if (t1y > t2y)
t[t11].y ++;
else if(t1y < t2y)
t[t11].s ++;
else
t[t11].p ++;
t[t22].yq += t2y;
t[t22].sq += t1y;
if (t1y > t2y)
t[t22].s ++;
else if(t1y < t2y)
t[t22].y ++;
else
t[t22].p ++;
}
for (int i = 0; i < tnum; i ++)
{
t[i].f = t[i].y * 3 + t[i].p;
t[i].z = t[i].y + t[i].p + t[i].s;
t[i].zq = t[i].yq - t[i].sq;
}
sort(t, t + tnum, cmp);
printf("%s\n", cn);
for (int i = 0; i < tnum; i ++)
printf("%d) %s %dp, %dg (%d-%d-%d), %dgd (%d-%d)\n",i + 1, t[i].name, t[i].f, t[i].z, t[i].y, t[i].p, t[i].s, t[i].zq, t[i].yq, t[i].sq);
if (tt)
printf("\n");
}
return 0;
}

UVA 10194 Football (aka Soccer)的更多相关文章

  1. D - Football (aka Soccer)

    Football the most popular sport in the world (americans insist to call it "Soccer", but we ...

  2. UVA 10194 (13.08.05)

    :W Problem A: Football (aka Soccer)  The Problem Football the most popular sport in the world (ameri ...

  3. UVA题目分类

    题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics ...

  4. Volume 1. Sorting/Searching(uva)

    340 - Master-Mind Hints /*读了老半天才把题读懂,读懂了题输出格式没注意,结果re了两次. 题意:先给一串数字S,然后每次给出对应相同数目的的一串数字Si,然后优先统计Si和S ...

  5. 刘汝佳 算法竞赛-入门经典 第二部分 算法篇 第五章 3(Sorting/Searching)

    第一题:340 - Master-Mind Hints UVA:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Item ...

  6. (Step1-500题)UVaOJ+算法竞赛入门经典+挑战编程+USACO

    http://www.cnblogs.com/sxiszero/p/3618737.html 下面给出的题目共计560道,去掉重复的也有近500题,作为ACMer Training Step1,用1年 ...

  7. ACM训练计划step 1 [非原创]

    (Step1-500题)UVaOJ+算法竞赛入门经典+挑战编程+USACO 下面给出的题目共计560道,去掉重复的也有近500题,作为ACMer Training Step1,用1年到1年半年时间完成 ...

  8. 算法竞赛入门经典+挑战编程+USACO

    下面给出的题目共计560道,去掉重复的也有近500题,作为ACMer Training Step1,用1年到1年半年时间完成.打牢基础,厚积薄发. 一.UVaOJ http://uva.onlinej ...

  9. Randy Pausch’s Last Lecture

          he University of Virginia American Studies Program 2002-2003.                     Randy Pausch ...

随机推荐

  1. object 属性 对象的继承 (原型, call,apply)

    object 为一切对象的基类! 属性:constructor: 对创建对象的函数的引用Prototype: 原型(类型) hasOwnProperty(property):判断对象是否有某个特定的属 ...

  2. js传带参数的函数

    字符串: setTimeout('pageScroll(4)',100);

  3. php Static静态关键字

    静态属性与方法可以在不实例化类的情况下调用,直接使用类名::方法名的方式进行调用.静态属性不允许对象使用->操作符调用. class Car { private static $speed =  ...

  4. python使用sqlite3

    import sqlite3 mysqldb=sqlite3.connect(r"C:\Users\Administrator\Desktop\testdb.db") mysqld ...

  5. UITableView中复用cell显示信息错乱

    UITableView继承自UIScrollview,是苹果为我们封装好的一个基于scroll的控件.上面主要是一个个的 UITableViewCell,可以让UITableViewCell响应一些点 ...

  6. iOS 常用基础框架

    框架名称 功能 Foundation 提供OC的基础类(像NSObject).基本数据类型等 UIKit 创建和管理应用程序的用户界面 QuartzCore 提供动画特效以及通过硬件进行渲染的能力 C ...

  7. Ajax编程相对路径与绝对路径

    http://www.worlduc.com/blog2012.aspx?bid=16946309 ajax同一域名调用采用相对路径 var url = 'QuerySingleDataByField ...

  8. GIT在LINUX下的基本操作

    没办法,看来,VIM技能也要同步练起来了. 离开了WIN的日常应用安乐窝,外面的世界有多精彩? GIT的错了我再改..呵呵 git clone http://username@1.2.3.4/repo ...

  9. android中handler中 obtainmessge与New message区别

    obtainmessage()是从消息池中拿来一个msg 不需要另开辟空间new new需要重新申请,效率低,obtianmessage可以循环利用: //use Handler.obtainMess ...

  10. 解读30个提高Web程序执行效率的好经验

    其实微博是个好东西,关注一些技术博主之后,你不用再逛好多论坛了,因为一些很好的文章微博会告诉你,最近看到酷勤网推荐的一篇文章<30个提高Web程序执行效率的好经验>,文章写得不错,提到一些 ...