Tournament Chart【模拟+vector+map+string】
Tournament Chart
传送门:链接 来源:UPC10889
题目描述
In 21XX, an annual programming contest, Japan Algorithmist GrandPrix (JAG) has become one of the most popular mind sports events.
JAG is conducted as a knockout tournament. This year, N contestants will compete in JAG. A tournament chart is represented as a string. '[[a-b]-[c-d]]' is an easy example. In this case, there are 4 contestants named a, b, c, and d, and all matches are described as follows:
Match 1 is the match between a and b.
Match 2 is the match between c and d.
Match 3 is the match between [the winner of match 1] and [the winner of match 2].
More precisely, the tournament chart satisfies the following BNF:
<winner> ::= <person> | "[" <winner> "-" <winner> "]"
<person> ::= "a" | "b" | "c" | ... | "z"
You, the chairperson of JAG, are planning to announce the results of this year's JAG competition. However, you made a mistake and lost the results of all the matches. Fortunately, you found the tournament chart that was printed before all of the matches of the tournament. Of course, it does not contains results at all. Therefore, you asked every contestant for the number of wins in the tournament, and got N pieces of information in the form of "The contestant ai won vi times".
Now, your job is to determine whether all of these replies can be true.
输入
The input consists of a single test case in the format below.
S
a1 v1
:
aN vN
S represents the tournament chart. S satisfies the above BNF. The following N lines represent the information of the number of wins. The (i+1)-th line consists of a lowercase letter ai and a non-negative integer vi (vi≤26) separated by a space, and this means that the contestant ai won vi times. Note that N (2≤N≤26) means that the number of contestants and it can be identified by string S. You can assume that each letter ai is distinct. It is guaranteed that S contains each ai exactly once and doesn't contain any other lowercase letters.
输出
Print 'Yes' in one line if replies are all valid for the tournament chart. Otherwise, print 'No' in one line.
样例输入
[[m-y]-[a-o]]
o 0
a 1
y 2
m 0
样例输出
Yes
题目大意:
给出上面的BNF公式(看不懂),猜测应该是一个公式类似: [ [a-b]-c ] (这个比样例有普遍性)
每个 - 代表前面的队伍要和后面的队伍比赛,获胜的再与后面的比,在上面的例子中,是a先与b比赛,获胜者再与c比。
题目给出每个队伍最终的获胜次数,让你判断这个数据有没有可能是正确的。
解题思路:
按照下图一步步模拟就行了,因为要用到删除操作,直接用string不太好实现,可以用 vector<char>,这篇博很详细:传送门 。
具体实现就是每次遍历vector,找到两边都是字母的减号,删除【vc[i-2],vc[i+2]】的内容,再把胜场多的字符插入进去(也可以先修改字符再删内容,只是区间要变化一下),并将他对应的胜场-1(map数组在这很好用)。
判断条件:
两个比赛队伍,肯定会有一个会输,也就是胜场为0,否则就输出NO。
如果两个比赛的队伍胜场相同,也要输出NO。
最后只剩下的一个队伍对应的胜场应该也是0,否则输出NO。
AC代码:
感谢队友给找的bug:https://me.csdn.net/qq_43559193
#include<bits/stdc++.h>
using namespace std;
int main()
{
string a;
cin>>a; ///输入字符串
int la=a.size(),cnt=0;
for(int i=0;i<la;i++){
if(a[i]>='a'&&a[i]<='z'){
cnt++;
}
}
vector<char>vc;
for(int i=0;i<la;i++){ ///字符串内容赋值到vector中(输入的时候直接往里面存也可以)
vc.push_back(a[i]);
}
map<char,int>mp;
for(int i=0;i<cnt;i++){ ///输入胜场,存入mp数组
int num;
char id;
cin>>id>>num;
mp[id]=num;
}
while(vc.size()!=1){
for(int i=0;i<vc.size();i++){
if(vc[i]=='-'&&vc[i-1]>='a'&&vc[i-1]<='z'&&vc[i+1]>='a'&&vc[i+1]<='z'){
if(!(mp[vc[i+1]]==0||mp[vc[i-1]]==0)||(mp[vc[i+1]]==mp[vc[i-1]])){
cout<<"No"<<endl; ///一定要判断,不然会死循环
return 0;
}
if(mp[vc[i+1]]>mp[vc[i-1]]){ ///减号右边的胜场多
vc[i+2]=vc[i+1];///先修改值
mp[vc[i+1]]--;
vc.erase(vc.begin()+i-2,vc.begin()+i+2); ///删除区间内容
break;
}else if(mp[vc[i+1]]<mp[vc[i-1]]){///减号左边的胜场多
vc[i+2]=vc[i-1];
mp[vc[i-1]]--;
vc.erase(vc.begin()+i-2,vc.begin()+i+2);
break;
}
}
}
}
if(mp[vc[0]]==0) cout<<"Yes"<<endl;
else cout<<"No"<<endl;
return 0;
}
Tournament Chart【模拟+vector+map+string】的更多相关文章
- programming review (c++): (1)vector, linked list, stack, queue, map, string, bit manipulation
编程题常用知识点的review. most important: 想好(1)详尽步骤(2)边界特例,再开始写代码. I.vector #include <iostream> //0.头文件 ...
- POJ 3096 Surprising Strings(STL map string set vector)
题目:http://poj.org/problem?id=3096 题意:给定一个字符串S,从中找出所有有两个字符组成的子串,每当组成子串的字符之间隔着n字符时,如果没有相同的子串出现,则输出 &qu ...
- 2015-2016 ACM-ICPC, NEERC, Southern Subregional Contest A Email Aliases(模拟STL vector+map)
Email AliasesCrawling in process... Crawling failed Time Limit:2000MS Memory Limit:524288KB ...
- POJ 3087 Shuffle'm Up【模拟/map/string】
Shuffle'm Up Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 14471 Accepted: 6633 Descrip ...
- PAT 1039 Course List for Student (25分) 使用map<string, vector<int>>
题目 Zhejiang University has 40000 students and provides 2500 courses. Now given the student name list ...
- c++ list, vector, map, set 区别与用法比较
http://blog.csdn.net/alex_xhl/article/details/37692297 List封装了链表,Vector封装了数组, list和vector得最主要的区别在于ve ...
- UVA 156:Ananagrams (vector+map+sort)
题意:一大堆单词中间有空格隔开,以'#'结束输出,问只出现一次的的单词有哪些(如果两个具有相同的长度,相同的字母也算是相同的,不区分大小写,如:noel和lone属于一个单词出现两次).最后按照字典序 ...
- list, vector, map, set 区别与用法比较
List封装了链表,Vector封装了数组, list和vector得最主要的区别在于vector使用连续内存存储的,他支持[]运算符,而list是以链表形式实现的,不支持[]. Vector对于随机 ...
- 【数据结构】Tournament Chart
Tournament Chart 题目描述 In 21XX, an annual programming contest, Japan Algorithmist GrandPrix (JAG) has ...
随机推荐
- 洛谷P2468 粟粟的书架
题目链接:https://www.luogu.org/problemnew/show/P2468 知识点: 可持久化线段树.二分.前缀和 解题思路: 对于 \(R, C \le 200, M \le ...
- 启动独立的tomcat服务器,没有自动创建ServletContext,对Context生命周期的监听失败
1.可能web.xml文件里对ContextListener没有进行配置 2.web.xml文件有关对ContextListener的配置,出现了错误的单词拼写问题 比如 <listener&g ...
- C语言合法标识符(hud2024)
输入方式:先输入一个整型,再循环输入带空格的字符串. 思考:整型用scanf_s()输入.大循环输入字符串前用getchar()函数读取缓冲区的字符.然后,输入带空格的字符串就要用”gets_s()“ ...
- Robot Framework(1)- 入门介绍
如果你还想从头学起Robot Framework,可以看看这个系列的文章哦! https://www.cnblogs.com/poloyy/category/1770899.html RF 的介绍 R ...
- OAUTH开放授权
OAUTH开放授权 OAUTH开放授权为用户资源的授权提供了一个安全的.开放而又简易的标准.OAUTH的授权不会使第三方触及到用户的帐号信息例如用户名与密码等,即第三方无需使用用户的用户名与密码就可以 ...
- 使用EditPlus根据指定字符批量换行,快速填充Postman请求参数键值对
1.当某个.ext格式的文件中的重复格式内容太多时,而又想要根据某个字符进行批量换行时,那么可以使用EditPlus进行批量换行. 在开发过程中就会经常遇到这种问题,比如把Url的请求参数,快速的填写 ...
- Docker虚拟机配置手札(centos)
一.Docker只支持CentOS7及以上系统,不支持6.x系统 二.yum安装Docker 1.安装相关环境和设置仓库 yum install -y yum-utils device-mapper- ...
- 获取cookie里面的sessionid
做接口的时候我们需要获取一个令牌,类似于token以及sessionid这样的签名,也可以说是一个令牌. import urllib.request, urllib.parse, urllib.err ...
- python时间格式化、运行时长计算
1.格式化: import time timeStr=time.strftime('%Y-%m-%d %H:%M:%S',time.localtime()) #格式化为:2018-07-17 19:0 ...
- Rocket - debug - DMI
https://mp.weixin.qq.com/s/70BoeS7z4aBZK24zxdZzXA 简单介绍DMI的实现. 1. DMIConsts 定义DMI使用的常量: 其中: a. dmiDat ...