LEAGUE TABLES
时间限制: 1 Sec 内存限制: 128 MB
提交: 349 解决: 150
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题目描述
League football (known in some circles as soccer) has been played in England since 1888 and is the most popular winter game through most of Europe, just as rugby is in New Zealand. English newspapers and many Websites publish the latest results and the current tables.
With English football, each team plays every other team both home and away, and at the end of the season, the team with most points wins the title. Points are awarded for winning (3 points) or drawing (1 point each). Teams level on points are separated by the larger goal difference, that is goals for (ie scored) minus goals against (ie conceded). If this is level, the team who has scored more goals is placed first.
In this problem you will be given a list of teams and their current record, followed by a list of matches. You have to update the record of each team who has a result recorded and output a correctly sorted league table.
输入
The first line contains a single integer, T (8 <= T <= 30), which is the number of teams in the league. The next 2 x T lines each contain the current record of one team as follows:
The first line contains the team name which consists of one or more words, which may contain numbers. No name will be longer than 30 characters.
All numbers on the second line are non-negative integers.
The next line contains a single integer, G (0 < G <= (T/2)), which is the number of games recorded. There then follow 4 x G lines, each containing data on one game as follows:
The two teams will both be from the preceding list of teams. The scores will each be a non-negative integer less than 20.
输出
T lines giving the updated record of each team from the input, name followed by data all on 1 line. Teams are to be sorted by highest points, then highest goal difference, then most goals scored, then alphabetical order (case insensitive, numbers before letters).
样例输入
复制样例数据
18
Aston Villa
33 21 6 6 74 34 69
Sheffield United
34 18 12 4 63 33 66
Sunderland
33 18 3 12 47 34 57
Wolverhampton Wanderers
33 15 9 9 47 35 54
Derby County
33 13 8 12 43 42 47
Newcastle United
34 13 10 11 53 43 49
Manchester City
33 13 8 12 49 41 47
Nottingham Forest
34 13 8 13 56 55 47
Stoke City
34 13 8 13 37 45 47
Everton
33 13 7 13 46 30 46
Bury
33 13 6 14 40 57 45
Liverpool
33 13 5 15 47 45 44
Blackburn Rovers
33 12 4 17 47 60 40
West Bromwich Albion
34 11 8 15 43 51 41
Preston North End
33 12 4 17 37 45 40
Notts County
34 9 11 14 46 60 38
Burnley
34 11 5 18 34 54 38
Glossop
34 4 10 20 30 75 22
5
Aston Villa
3
Preston North End
1
Blackburn Rovers
2
Wolverhampton Wanderers
1
Derby County
2
Everton
1
Liverpool
2
Bury
0
Sunderland
3
Manchester City
1
样例输出
Aston Villa 34 22 6 6 77 35 72
Sheffield United 34 18 12 4 63 33 66
Sunderland 34 19 3 12 50 35 60
Wolverhampton Wanderers 34 15 9 10 48 37 54
Derby County 34 14 8 12 45 43 50
Newcastle United 34 13 10 11 53 43 49
Manchester City 34 13 8 13 50 44 47
Liverpool 34 14 5 15 49 45 47
Nottingham Forest 34 13 8 13 56 55 47
Stoke City 34 13 8 13 37 45 47
Everton 34 13 7 14 47 32 46
Bury 34 13 6 15 40 59 45
Blackburn Rovers 34 13 4 17 49 61 43
West Bromwich Albion 34 11 8 15 43 51 41
Preston North End 34 12 4 18 38 48 40
Notts County 34 9 11 14 46 60 38
Burnley 34 11 5 18 34 54 38
Glossop 34 4 10 20 30 75 22
题目大意：
先输入一个整数nnn，代表有nnn支队伍，下面2n2n2n行，每两行代表一支队伍当前的信息，先输入队伍的名称，由字母、数字、空格组成，然后第二行依次输入当前的比赛场数、胜场数、平局数、败场数、进球数、失球数、得分。然后输入一个数字ttt，代表还需进行ttt场比赛，下面4t4t4t行代表比赛的信息，每场比赛先输入主场方的队伍名称，再输入该场比赛的进球数，然后输入客场队伍的名称，其队伍的进球数。最后输出ttt场比赛打完后各个队伍的所有信息，先按得分排序，得分高得优先，得分相同者，净球数（进球数-失球数）大者优先，净球数相同者，进球数多者优先，进球数相同者，队伍名字典序小者优先。
解题思路：
此题只需模拟一下即可，需注意的是，在比较队伍名字典序时，需将队伍名中的字母提取出来，不区分大小下来进行比较。
代码：
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>
#include <utility>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define inf 0x3f3f3f3f
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define lep(i,l,r) for(int i=l;i>=r;i--)
#define ms(arr) memset(arr,0,sizeof(arr))
//priority_queue<int,vector<int> ,greater<int> >q;
const int maxn = (int)1e5 + 5;
const ll mod = 1e9+7;
struct node
{string name;string str;int num;int win;int draw;int lose;int jin;int shi;int point;int cha;
}arr[120];
map<string,int> m;
bool cmp(node a,node b) {if(a.point==b.point) {if(a.cha==b.cha) {if(a.jin==b.jin) {return a.str<b.str;}else return a.jin>b.jin;}else return a.cha>b.cha;}else return a.point>b.point;
}
int main()
{#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);#endif//freopen("out.txt", "w", stdout);//ios::sync_with_stdio(0),cin.tie(0);int n;cin>>n;rep(i,1,n) {while(getchar()!='\n');getline(cin,arr[i].name);m[arr[i].name]=i;arr[i].str="";for(int j=0;j<arr[i].name.size();j++) {if(arr[i].name[j]>='a'&&arr[i].name[j]<='z')arr[i].str+=arr[i].name[j];if(arr[i].name[j]>='A'&&arr[i].name[j]<='Z') arr[i].str+=(arr[i].name[j]+32);}cin>>arr[i].num>>arr[i].win>>arr[i].draw>>arr[i].lose>>arr[i].jin>>arr[i].shi>>arr[i].point;arr[i].cha=arr[i].jin-arr[i].shi;}int t;cin>>t;string s1,s2;int a,b;rep(i,1,t) {while(getchar()!='\n');getline(cin,s1);cin>>a;while(getchar()!='\n');getline(cin,s2);cin>>b;arr[m[s1]].num++;arr[m[s2]].num++;if(a==b) {arr[m[s1]].draw++;arr[m[s2]].draw++;arr[m[s1]].jin+=a;arr[m[s1]].shi+=b;arr[m[s2]].jin+=b;arr[m[s2]].shi+=a;arr[m[s1]].point++;arr[m[s2]].point++;arr[m[s1]].cha=arr[m[s1]].jin-arr[m[s1]].shi;arr[m[s2]].cha=arr[m[s2]].jin-arr[m[s2]].shi;}else if(a>b) {arr[m[s1]].win++;arr[m[s2]].lose++;arr[m[s1]].jin+=a;arr[m[s1]].shi+=b;arr[m[s2]].jin+=b;arr[m[s2]].shi+=a;arr[m[s1]].point+=3;arr[m[s1]].cha=arr[m[s1]].jin-arr[m[s1]].shi;arr[m[s2]].cha=arr[m[s2]].jin-arr[m[s2]].shi;}else {arr[m[s1]].lose++;arr[m[s2]].win++;arr[m[s1]].jin+=a;arr[m[s1]].shi+=b;arr[m[s2]].jin+=b;arr[m[s2]].shi+=a;arr[m[s2]].point+=3;arr[m[s1]].cha=arr[m[s1]].jin-arr[m[s1]].shi;arr[m[s2]].cha=arr[m[s2]].jin-arr[m[s2]].shi;}}sort(arr+1,arr+1+n,cmp);rep(i,1,n) {cout<<arr[i].name<<" "<<arr[i].num<<" "<<arr[i].win<<" "<<arr[i].draw<<" "<<arr[i].lose<<" "<<arr[i].jin<<" "<<arr[i].shi<<" "<<arr[i].point<<endl;}return 0;
}
