这个题自己感觉就是结构体排序的问题 ，不过细节问题好像很多 ，，而且还得把题读得十分明白 ，这种超长的题目，果然还要锻炼自己英语功底 ，那几个足球术语都是海词现查的
今天（应该是昨天了） 没写完 明天（也就是今天） 继续把这个弄完 STL用的不好问题很多
| Problem A: Football (aka Soccer) |
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):
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.
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:
There must be a single blank space between fields and a single blank line between output sets. See the sample output for examples.
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
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)
? 2001 Universidade do Brasil (UFRJ). Internal Contest 2001.
1 /**UVAOJ 10194 对多个项目进行排序**/ 2 #include<iostream> 3 #include<stdio.h> 4 #include<set> 5 #include<string> 6 #include<algorithm> 7 using namespace std; 8 9 typedef struct team 10 { 11 string name; 12 //int rank=0; 13 int points=0; 14 int played=0; 15 int wins=0; 16 int ties = 0; 17 int lose = 0; //输的比赛次数 18 int losses = 0; //输球几个 19 int differ = 0; 20 int goals = 0; 21 22 }team; 23 24 bool cmp(const team *a, const team*b) //自定义比较函数 25 { 26 if (a->points > b->points) return true; 27 else if (a->points < b->points) return false; 28 else 29 { 30 if (a->wins>b->wins) return true; 31 else if (a->wins < b->wins) return false; 32 else 33 { 34 if (a->differ>b->differ) return true; 35 else if (a->differ < b->differ) return false; 36 else 37 { 38 if (a->goals>b->goals) return true; 39 else if (a->goals < b->goals) return false; 40 else 41 { 42 if (a->played<b->played) return true; 43 else if (a->played >= b->played) return false; 44 } 45 } 46 } 47 } 48 } 49 50 int main(void) 51 { 52 string game; 53 int matches; 54 int num; 55 scanf("%d", &matches); 56 getchar(); 57 for (int i = 0; i < matches; i++) 58 { 59 getline(cin, game); 60 scanf("%d", &num); 61 set<team*> T; 62 for (int oo = 0; oo < num; oo++) 63 { 64 string teamtmp; 65 team *tmp=(team*)malloc(sizeof(team)); //是不是应该用内存分配？ 66 cin >> teamtmp; 67 tmp->name = teamtmp; 68 T.insert(tmp); 69 } //存入各个小组信息 70 scanf("%d", &num); //比赛场数 71 for (int oo = 0; oo < num; oo++) 72 { 73 string data; //读入信息的区域 74 getline(cin, data); 75 string cur; 76 int cnt = 0; 77 team *curteama,*curteamb; 78 int scorea = 0 , scoreb = 0; 79 while (data[cnt] != ‘#‘) cur += data[cnt++]; 80 for (set<team*>::iterator i = T.begin(); i != T.end(); i++) 81 if ((*i)->name == cur) curteama = *i; //定位到当前队伍a 82 while (data[cnt] != ‘@‘) //读取比分 先读 a team 83 { 84 scorea = scorea * 10 + data[cnt] - ‘0‘; 85 cnt++; 86 } 87 cnt++; 88 while (data[cnt] != ‘#‘) 89 { 90 scoreb = scoreb * 10 + data[cnt] - ‘0‘; //再读b team 91 cnt++; 92 } 93 while (data[cnt] != ‘\0‘) cur += data[cnt++]; 94 for (set<team*>::iterator i = T.begin(); i != T.end(); i++) //定位到当前队伍b 95 if ((*i)->name == cur) curteamb = *i; 96 //开始判断 并记录 97 curteama->played += scorea; 98 curteamb->played += scoreb; 99 curteama->losses += scoreb; 100 curteamb->losses += scorea; 101 curteama->goals += scorea; 102 curteamb->goals += scoreb; 103 curteama->played++; 104 curteamb->played++; 105 if (scorea > scoreb) //分为a赢了b a b平 a输给 b三种情况 106 { 107 curteama->wins++; 108 curteamb->lose++; 109 curteama->points += 3; 110 curteamb->points += 0; 111 } 112 else if (scorea < scoreb) 113 { 114 curteamb->wins++; 115 curteama->lose++; 116 curteamb->points += 3; 117 curteama->points += 0; 118 } 119 else if (scorea == scoreb) 120 { 121 curteama->ties++; 122 curteamb->ties++; 123 curteama->points++; 124 curteamb->points++; 125 } 126 //判断结束 127 for (set<team*>::iterator i = T.begin(); i != T.end(); i++) 128 { 129 (*i)->differ = (*i)->goals - (*i)->losses; 130 } 131 //sort(T.begin(), T.end(), cmp); 132 int count = 1; 133 cout << game << endl; 134 for (set<team*>::iterator i = T.begin(); i != T.end(); i++) 135 { 136 cout << count << ") " << (*i)->name << " " << (*i)->points << "p, " << (*i)->played << "g (" << (*i)->wins << "-" << (*i)->ties << "-" << (*i)->lose << "), " << (*i)->differ << "gd (" << (*i)->goals << "-" << (*i)->losses << ")" << endl; 137 } 138 printf("\n"); 139 } 140 } 141 getchar(); 142 return 0; 143 }
UVA10194 FootBall aka Soccer,布布扣,bubuko.com
原文：http://www.cnblogs.com/VOID-133/p/3570507.html