【map+审题】#19 A. World Football Cup

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#map #codeforces

A. World Football Cup
time limit per test
2 seconds
memory limit per test
64 megabytes
input
standard input
output
standard output

Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations:

  • the final tournament features n teams (n is always even)
  • the first n / 2 teams (according to the standings) come through to the knockout stage
  • the standings are made on the following principle: for a victory a team gets 3 points, for a draw — 1 point, for a defeat — 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place — in decreasing order of the difference between scored and missed goals; in the third place — in the decreasing order of scored goals
  • it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity.

You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.

Input

The first input line contains the only integer n (1 ≤ n ≤ 50) — amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following n·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, wherename1, name2 — names of the teams; num1, num2 (0 ≤ num1, num2 ≤ 100) — amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.

Output

Output n / 2 lines — names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.

Sample test(s)
input 
4
A
B
C
D
A-B 1:1
A-C 2:2
A-D 1:0
B-C 1:0
B-D 0:3
C-D 0:3
output 
A
D
input 
2
a
A
a-A 2:1
output 
a


———————————————————————————————————————————————

《Updated 2015/07/25》

愚蠢的我哟，这是来自一年后的我对一年前的我伸出援手：

map<typeA,typeB>中也是可以使用结构体的~

为了既可以使用map的插入优势，又想要使用数组排序，那就先用map建造一个 map<string,team>（其中string为名字，team为包括名字的结构体，包括名字是为了最终提取出来的时候依然包含名字）；在map插入完毕之后，使用迭代器一个一个提取出来放入数组，对数组进行cmp排序即可。

这题的难(keng)点依然在于要好好审题，具体点就是：

先按照能力排序（看得分，相同时看净胜球，相同时看进球数）获得前一半的队伍

将这些队伍按照字典升序输出，AC代码如下：

C++ Source Code Below:

#include <map>
#include <cmath> 
#include <cctype>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))

struct team
{
	int w,l;
	int score;
	string name;
	team(){w=0,l=0,score=0;}
} t[64];

map<string,team> mst;

bool cmp(const team &a, const team &b)
{
	if(a.score==b.score)
		if( (a.w-a.l)==(b.w-b.l) )
			return a.w>b.w;			
		else return (a.w-a.l)>(b.w-b.l);	
	else return a.score>b.score;
}

bool scmp(const team &a, const team &b)
{
	return a.name<b.name;
}

int main()
{
	int n=0;
	scanf("%d",&n);
	mst.clear();
	for(int i=0;i<n;i++)
	{
		cin>>t[i].name;
		mst[t[i].name].name=t[i].name;	
	} 
	int a=0,b=0;
	string battle,t1,t2;
	for(int i=0;i<(n-1)*n/2;i++)
	{	
		cin>>battle; 
		scanf("%d:%d",&a,&b);
		int div = battle.find('-');	//pos of '-'
		t1=battle.substr(0,div);
		t2=battle.substr(div+1,battle.length()-div-1);
		mst[t1].w+=a; mst[t2].w+=b;
		mst[t1].l+=b; mst[t2].l+=a;
		if(a==b)
		{
			mst[t1].score++;
			mst[t2].score++;
		}
		else if(a>b) mst[t1].score+=3;
		else mst[t2].score+=3;
	}
	int tpos=0;
	for(map<string,team>::iterator mit=mst.begin();mit!=mst.end();++mit)
	{
		t[tpos++] = mit->second;
		//cout<<t[tpos-1].name<<endl;
	}
	sort(t,t+n,cmp);
	sort(t,t+n/2,scmp);
	for(int i=0;i<n/2;i++) cout<<t[i].name<<endl;
	return 0;
}


———————————————————————————————————————————————


这次试了试map，map呢是一种Key&Value的形式，其特点是会随时按照Key排序（顺序），这一点对维护来说非常棒，但是Key是唯一固定的，不可以修改，只能删除。

这道题如果不需要最终将晋级的队伍按字典序输出的话，如下代码就是正确的：（作为初级map模板）

#include<map> //Unique Key,Default Sort by key
#include<cstdio>
#include<string>
#include<cstdlib> 
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

//defination
typedef pair<string,int> si;
typedef map<string,int> t_map;
typedef map<int,string> a_map;


int cmp(const si& x,const si& y){return x.second>y.second;}

int main()
{
	int n=0;
	scanf("%d",&n);
	t_map team;
	a_map team_ans;
	//memset(team,0,sizeof(team));
	team.clear();
	team_ans.clear();
	for(int i=0;i<n;i++) //Read all team names by map
	{
		string name;
		cin>>name;
		//cout<<name;
		team.insert(pair<string,int>(name,0));
		//OR  insert(make_pair<string,int>(name,0));
		//OR  team[name]= 0;
		//Hint: Key can not be changed when Value can
	}
	for(int j=1;j<=(n-1)*n/2;j++) //get their goals and points
	{
		t_map::iterator it1,it2;
		char na1[31],na2[31],huiche;
		int num1,num2,flag=0,k=0;
		int cnt=0;
		//scanf("%s %d:%d",names,&num1,&num2);
		scanf("%c",&huiche);
		while(1)	//1st team
		{
			cnt=0;
			char tmp;
			scanf("%c",&tmp);
			if(tmp=='-')break;
			na1[cnt++]=tmp;
		}
		na1[cnt+1]='\0';
		while(1)	//2nd team
		{
			cnt=0;
			char tmp;
			scanf("%c",&tmp);
			if(tmp==' ')break;
			na2[cnt++]=tmp;
		}
		na2[cnt+1]='\0';
		string name1(na1);
		string name2(na2);
		//cout<<"1:"<<name1<<endl<<"2:"<<name2<<endl; 
		int plus1=0,plus2=0;
		scanf("%d:%d",&num1,&num2);
		if(num1>num2)plus1=3,plus2=0;
		else if(num1<num2)plus1=0,plus2=3;
		else plus1=plus2=1;
		team[name1] -= plus1;//According to default sort  ,but we need reverse,so
		team[name2] -= plus2;//According to default sort  ,but we need reverse,so
		//cout<<"1:"<<team[name1]<<endl<<"2:"<<team[name2]<<endl; 
	}
	for(t_map::iterator iter=team.begin();iter!=team.end();++iter)
	{
		team_ans.insert(make_pair<int,string>(iter->second,iter->first));
	}
	int cnt=0;
	for(a_map::iterator iter2=team_ans.begin();iter2!=team_ans.end();++iter2)
	{
		if(cnt==0)cout<<iter2->second;
		else cout<<endl<<iter2->second;
		cnt++;
		if(cnt==n/2)break;
	}
	return 0;
}

但这个是需要字典序的真讨厌……于是我作死的又转了一遍回来： 

#include<map> //Unique Key,Default Sort by key
#include<cstdio>
#include<string>
#include<cstdlib> 
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

//defination
typedef pair<string,int> si;
typedef map<string,int> t_map;
typedef map<int,string> a_map;

char ans[26][32]; 
int cmp(char* &a,char* &b){return strcmp(a,b);}
int cmp(const si& x,const si& y){return x.second>y.second;}

int main()
{
	int n=0;
	scanf("%d",&n);
	t_map team,team_fin;
	a_map team_ans;
	team.clear();
	team_ans.clear();
	team_fin.clear();
	for(int i=0;i<n;i++) //Read all team names by map
	{
		string name;
		cin>>name;
		//cout<<name;
		team.insert(pair<string,int>(name,0));
		//OR  insert(make_pair<string,int>(name,0));
		//OR  team[name]= 0;
		//Hint: Key can not be changed when Value can
	}
	for(int j=1;j<=(n-1)*n/2;j++) //get their goals and points
	{
		t_map::iterator it1,it2;
		char na1[31],na2[31],huiche;
		int num1,num2,flag=0,k=0;
		int cnt=0;
		//scanf("%s %d:%d",names,&num1,&num2);
		scanf("%c",&huiche);
		cnt=0;
		while(1)	//1st team
		{
			char tmp;
			scanf("%c",&tmp);
			if(tmp=='-')break;
			na1[cnt++]=tmp;
		}
		na1[cnt]='\0';
		cnt=0;// WA- This must be here ,not in the while(1)
		while(1)	//2nd team
		{
			char tmp;
			scanf("%c",&tmp);
			if(tmp==' ')break;
			na2[cnt++]=tmp;
		}
		na2[cnt]='\0';
		string name1(na1);
		string name2(na2);
		//cout<<"1:"<<name1<<endl<<"2:"<<name2<<endl; 
		int plus1=0,plus2=0;
		scanf("%d:%d",&num1,&num2);
		if(num1>num2)plus1=3,plus2=0;
		else if(num1<num2)plus1=0,plus2=3;
		else plus1=plus2=1;
		team[name1] -= plus1;//According to default sort  ,but we need reverse,so
		team[name2] -= plus2;//According to default sort  ,but we need reverse,so
		//cout<<"1:"<<team[name1]<<endl<<"2:"<<team[name2]<<endl; 
	}
	for(t_map::iterator iter=team.begin();iter!=team.end();++iter)//Change Key and Value in another map
	{
		team_ans.insert(make_pair<int,string>(iter->second,iter->first));
	}
	int cnt=0;
	for(a_map::iterator iter2=team_ans.begin();iter2!=team_ans.end();++iter2)//Change back to get a half size answer map
	{
		//if(cnt==0)cout<<iter2->second;
		//else cout<<endl<<iter2->second;
		team_fin.insert(make_pair<string,int>(iter2->second,iter2->first));
		cnt++;
		if(cnt==n/2)break;
	}
	for(t_map::iterator iter3=team_fin.begin();iter3!=team_fin.end();++iter3)
	{
		cout<<iter3->first<<endl;
	}
	
	return 0;
}

居然……WA at 5了……

这道题网上只有一个解题报告，而且跟我的好不一样，虽然觉得烦但是也没有退路了……加油debug吧Q^Q

虽说……怎么看怎么是对的哇…… 到底哪里出问题了哇……

看了看题意：我！了！个！去！

如果得分相同就看 (进球数 - 失球数) 的值, 如果还是相同就看进球数 (都是大的排在前面)

得分》净胜球》进球数

输出时：字典序。

长题目一定要好好审题Q^Q

解题报告 From:yunyouxi0@优快云

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <set>
#include <algorithm>

using namespace std;

char str[100][100];
char s[100];
char s1[100], s2[100];
struct P {
  string name;
  int s;
  int w, l;
}p[110];

bool cmp(const P & a, const P & b ) {
  if(a.s == b.s) {
    if(a.w - a.l == b.w - b.l) {
      return a.w > b.w;
    }
    else return a.w - a.l > b.w - b.l;
  }
  return a.s > b.s;
}

bool cmp2(const P & a, const P & b) {
  return a.name < b.name;
}

int main() {
  int n, a, b;
  while(scanf("%d", &n) != EOF) {
    map<string,int>mm;
    map<string,int>w;
    map<string,int>l;
    set<string>name;
    set<string>::iterator ps;
    for( int i=0; i<n; i++ ) {
      scanf("%s", str[i]);
    }
    for( int i=0; i<n*(n-1)/2; i++ ) {
      scanf("%s", s);
      int len = strlen(s), k1 = 0, k2 = 0;
      bool f = 0;
      for( int i=0; i<len; i++ ) {
        if(s[i] == '-') {
          f = 1;
          continue;
        }
        if(f) s2[k2++] = s[i];
        else s1[k1++] = s[i];
      }
      s1[k1] = '\0', s2[k2] = '\0';
      scanf("%d:%d", &a, &b);
      string ts1 = s1;
      string ts2 = s2;
      if(name.find(ts1) == name.end()) name.insert(ts1);
      if(name.find(ts2) == name.end()) name.insert(ts2);
      w[ts1] += a; l[ts1] += b;
      w[ts2] += b; l[ts2] += a;
      if(a > b) mm[ts1] += 3;
      else if(a == b) mm[ts1]++, mm[ts2]++;
      else mm[ts2] += 3;
    }
    int k = 0;
    for( ps=name.begin(); ps != name.end(); ps++ ) {
      p[k].name = *ps;
      p[k].s = mm[*ps];
      p[k].w = w[*ps];
      p[k].l = l[*ps];
      k++;
    }
    sort(p, p + k, cmp);
    sort(p, p + n/2, cmp2);
    for( int i=0; i<n/2; i++ ) cout<<p[i].name<<endl;
  }
  return 0;
}



话说刚看到个解题报告： From：UsingTC @sina

解法:直接模拟

吾辈发现……这是个CE的代码呀……而且……感觉……有点……乱，稍微改了改（测试这里的bug都交了5发WA）：

算了不贴了，弃掉。


Python强人真多：粘贴备看

import os
import sys

# solve
def solve():
    n = int(raw_input())
    dic = {}
    for i in range(n):
        str = raw_input()
        dic[str] = [0,0,0]
    
    i = n*(n-1)/2
    while i:
        list = raw_input().split()
        tmp1 = list[0].split("-")
        tmp2 = list[1].split(":")
        i -= 1

        if(int(tmp2[0]) > int(tmp2[1])):
            dic[tmp1[0]][0] += 3
            dic[tmp1[0]][1] += int(tmp2[1])
            dic[tmp1[0]][2] += int(tmp2[0])
            
            dic[tmp1[1]][1] += int(tmp2[0])
            dic[tmp1[1]][2] += int(tmp2[1])
        elif(int(tmp2[0]) == int(tmp2[1])):
            dic[tmp1[0]][0] += 1
            dic[tmp1[0]][1] += int(tmp2[1])
            dic[tmp1[0]][2] += int(tmp2[0])

            dic[tmp1[1]][0] += 1
            dic[tmp1[1]][1] += int(tmp2[0])
            dic[tmp1[1]][2] += int(tmp2[1])
        else:
            dic[tmp1[0]][1] += int(tmp2[1])
            dic[tmp1[0]][2] += int(tmp2[0])

            dic[tmp1[1]][0] += 3
            dic[tmp1[1]][1] += int(tmp2[0])
            dic[tmp1[1]][2] += int(tmp2[1])

    # out
    ans = sorted(dic.items() , key = lambda value:(value[1][0],value[1][2]-value[1][1],value[1][2]) , reverse = True)
    out = []
    for i in range(n/2):
        out.append(ans[i][0])
    out.sort()
    for s in out:
        print s

# main
if __name__ == "__main__":
    solve()



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题目地址：UVALive 7147 题意：一场比赛，赢的队得A分，输的得C分，平手都得B分。有n支队伍，分别比赛一次，选前m个队伍晋级（分数相同的随机排名）。问：可能的最大淘汰分数，可能的最小晋级分数。#include <cstdio> #include <cstring> #include <cmath> #include <map> #include <set> #include <vecto
19A World Football Cup
memory_qianxiao的博客
12-16 499
A. World Football Cup time limit per test 2 seconds memory limit per test 64 megabytes input standard input output standard output Everyone knows that 2010 FIFA World Cu
CF1215A Yellow Cards 水题 细节
Focus5679的博客
09-18 530
A. Yellow Cards time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The final match of the Berland Football Cup has been held recently...
A. Yellow Cards ( Codeforces Round #585 (Div. 2) 思维水题
weixin_30819085的博客
09-17 277
---恢复内容开始--- output standard output The final match of the Berland Football Cup has been held recently. The referee has shownnyellow cards throughout the match. At the beginning of t...
CF1363C Game On Leaves 题解
ShineEternal的笔记小屋
06-06 343
description： 给定 nnn 个节点的无根树。 两名选手轮流操作，每次可以选择一个叶节点并删除它以及所有和它相连的边。叶节点指度数不超过 111 的节点。删除节点 xxx 的选手胜利。 你需要判断先手是否有必胜策略。 solution： 首先观察到这道题的一个特判点：如果 xxx 刚开始就在叶子节点上，那么先手必然胜利。 注意不要忘了一个小 feature：...
584. Find Customer Referee
weixin_39436195的博客
07-26 965
Question Given a table customer holding customers information and the referee. +------+------+-----------+ | id | name | referee_id| +------+------+-----------+ | 1 | Will | NULL | |
Codeforces 1215
weixin_30271335的博客
09-20 177
A. 解 分类讨论。 Code #include<bits/stdc++.h> using namespace std; const int maxn=100003; int main(){ int a1,a2,k1,k2,n; cin>>a1>>a2>>k1>>k2>>n; if(k1&gt...
Codeforces 1215 A. Yellow Cards
邵光亮的博客
03-30 14万+
题意： 两支球队，111 队有 a1a_1a1​ 人，如果拿到 k1k_1k1​ 个黄牌就下场， 222 队有 a2a_2a2​ 人，如果拿到 k2k_2k2​ 个黄牌就下场。裁判有 nnn 张黄牌，问这场比赛最少和最多会有多少个人下场。 最少罚下场的人：111 队每人吃 (k1−1)(k_1-1)(k1​−1) 张黄牌, 222 队每人吃 (k2−1)(k_2-1)(k2​−1) 张黄牌，剩...
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