标签:queue ddr The std table inpu getch oppo ons
Problem description
There are n employees in Alternative Cake Manufacturing (ACM). They are now voting on some very important question and the leading world media are trying to predict the outcome of the vote.
Each of the employees belongs to one of two fractions: depublicans or remocrats, and these two fractions have opposite opinions on what should be the outcome of the vote. The voting procedure is rather complicated:
You know the order employees are going to vote and that they behave optimal (and they also know the order and who belongs to which fraction). Predict the outcome of the vote.
Input
The first line of the input contains a single integer n (1?≤?n?≤?200?000) — the number of employees.
The next line contains n characters. The i-th character is ‘D‘ if the i-th employee is from depublicans fraction or ‘R‘ if he is from remocrats.
Output
Print ‘D‘ if the outcome of the vote will be suitable for depublicans and ‘R‘ if remocrats will win.
Examples
5
DDRRR
D
6
DDRRRR
R
Note
Consider one of the voting scenarios for the first sample:
解题思路:仔细读一下题目,还是挺简单的。就是有两个门派D和R,给定一个字符串(只由D和R组成),只要位置靠前的人就可以将后面的对手deny掉,当轮到这个对手时,由于已被前面的人deny掉,所以此人再也没有deny他人的权利,直接跳过。不断循环,后面的人也可以deny前面还有选择权利的人,怎么实现呢?我们用队列来维护它们的位置(下标),当前面的人deny后面的对手后,此时同时出两个队的队首元素(表示位置上的人已经失效),但deny别人的人可能会被后面的对手deny掉,于是只需将其位置序号加上n后入自己的队列中(满足位置序号比后面大,即后面的人可以deny掉前面还有选择权的对手)。最终,哪个队列不为空,谁就拥有vote的权利。
AC代码:
1 #include<bits/stdc++.h> 2 using namespace std; 3 int n;char s[200005]; 4 queue<int> d,r; 5 int main(){ 6 cin>>n;getchar(); 7 cin>>s; 8 for(int i=0;i<n;++i){ 9 if(s[i]==‘D‘)d.push(i); 10 else r.push(i); 11 } 12 while(!d.empty()&&!r.empty()){ 13 int dd=d.front(),rr=r.front(); 14 if(dd<rr){d.pop();r.pop();d.push(n+dd);} 15 else{d.pop();r.pop();r.push(n+rr);} 16 } 17 if(!d.empty())cout<<‘D‘<<endl; 18 else cout<<‘R‘<<endl; 19 return 0; 20 }
标签:queue ddr The std table inpu getch oppo ons
原文地址:https://www.cnblogs.com/acgoto/p/9123141.html