Vacant Seat（Atcoder-C-交互式题目）

Posted 2020-10-23 CountingStars

C - Vacant Seat

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Time limit : 2sec / Memory limit : 256MB

Score : 500 points

Problem Statement

This is an interactive task.

Let N be an odd number at least 3.

There are N seats arranged in a circle. The seats are numbered 0 through N−1. For each i (0≤i≤N−2), Seat i and Seat i+1 are adjacent. Also, Seat N−1 and Seat 0 are adjacent.

Each seat is either vacant, or oppupied by a man or a woman. However, no two adjacent seats are occupied by two people of the same sex. It can be shown that there is at least one empty seat where N is an odd number at least 3.

You are given N, but the states of the seats are not given. Your objective is to correctly guess the ID number of any one of the empty seats. To do so, you can repeatedly send the following query:

• Choose an integer i (0≤i≤N−1). If Seat i is empty, the problem is solved. Otherwise, you are notified of the sex of the person in Seat i.

Guess the ID number of an empty seat by sending at most 20 queries.

Constraints

• N is an odd number.
• 3≤N≤99 999

Input and Output

First, N is given from Standard Input in the following format:

N

Then, you should send queries. A query should be printed to Standart Output in the following format. Print a newline at the end.

i

The response to the query is given from Standard Input in the following format:

s

Here, s is Vacant, Male or Female. Each of these means that Seat i is empty, occupied by a man and occupied by a woman, respectively.

Notice

• Flush Standard Output each time you print something. Failure to do so may result in TLE.
• Immediately terminate the program when s is Vacant. Otherwise, the verdict is indeterminate.
• The verdict is indeterminate if more than 20 queries or ill-formatted queries are sent.

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Sample Input / Output 1

In this sample, N=3, and Seat 0, 1, 2 are occupied by a man, occupied by a woman and vacant, respectively.

 

Input	Output
3
	0
Male
	1
Female
	2
Vacant

 

 

题意：给你大于等于3的奇数个座位，编号0到N-1,座位情况为空，男，女，其中同性不相邻，至少存在一个空座位，然后你询问一个座位编号，它告诉你现在那个地方是不是空，时空就结束，如果不是就告诉你这个地方的人的性别。要在20次查询以内得出结果！所以必然需要二分！

分析：这是第一次看到交互式题目，真的是看的我很懵逼！觉得很有必要写个博客！交互式，和我们原来做题目的模式是刚好相反的！像题目中给出的案例：输入N以后，输出0，然后我们输入Male，代表0号位置上面坐的男性，下面同样！但是以往我们做的非交互题目都是输入座位编号，然后电脑输出座位的信息！

（注意：这一所谓的性别相同的不能相邻是说中间假设存在空座位的两个性别相同的人之间是相邻的！或者直接相邻！

举个例子：5和7

N=5时，假设0位置为男性，mid=2，如果2号位置与0号位置同性别（假设为男），则一号位置必为女，所以去另外一边查找，left=mid；在left和mid位置同性别的情况下，只要mid-left为偶数就意味着中间不存在空座位；如果left与mid位置性别不同，即本例中的0号和2号位置的性别不同，那么他们中间必然为空；

N=7时，同上假设，mid=3，如果0号和3号位置同性别，就意味着中间两个必然有一个为空，所以right=mid，如果性别不同就意味着中间必然是两种性别交替出现，必然不存在空座位，所以此时left=mid，要注意此时的性别也要换成mid位置的性别。可以画图认真体会！

AC代码：

 

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<stack>
#include<map>
#include<vector>
#include<queue>
#define N 15
using namespace std;
typedef long long ll;
char s[N];
int main()
{
    ios::sync_with_stdio(false);
    int n,left,right,mid;
    int M1,M2;//用于标记性别
    cin>>n;
    cout << "0" << endl;
    cin>>s;
    if (s[0]==‘V‘)
       return 0;
    if (s[0]==‘M‘)  M1=1;
    else  M1=0;
    left=0,right=n;
    while (left<right)
    {
        mid=(right+left)/2;
        cout << mid << endl;
        cin>>s;
        if (s[0]==‘V‘)
          return 0;
        if (s[0]==‘M‘)  M2=1;
        else  M2=0;
        int t=(mid-left);
        if (t%2==0)
        {
            if (M1==M2) left=mid;
            else right=mid;
        }
        else
        {
            if (M1==M2) right=mid;
            else left=mid,M1=M2;
        }
    }
    cout << flush;
    return 0;
}