GYM 100685 J【交互题】

痛定思痛。 2022-08-04 14:40 130阅读 0赞

俄罗斯的人经常出一些交互题，比如强制离线之类的题目  
这题是二分+交互  
对于每一盏灯i，我们假设前面的灯位置都排好了位置，那么就二分那些这一盏灯所在的位置，询问的次数是nlog(n)次。  
另外如果死循环的话，那就是没有方案，设置一个cnt上限来判断死循环。

// whn6325689
    // Mr.Phoebe
    // http://blog.csdn.net/u013007900
    #include <algorithm>
    #include <iostream>
    #include <iomanip>
    #include <cstring>
    #include <climits>
    #include <complex>
    #include <fstream>
    #include <cassert>
    #include <cstdio>
    #include <bitset>
    #include <vector>
    #include <deque>
    #include <queue>
    #include <stack>
    #include <ctime>
    #include <set>
    #include <map>
    #include <cmath>
    #include <functional>
    #include <numeric>
    #pragma comment(linker, "/STACK:1024000000,1024000000")
    
    
    using namespace std;
    
    #define eps 1e-9
    #define PI acos(-1.0)
    #define INF 0x3f3f3f3f
    #define LLINF 1LL<<62
    #define speed std::ios::sync_with_stdio(false);
    
    typedef long long ll;
    typedef unsigned long long ull;
    typedef long double ld;
    typedef pair<ll, ll> pll;
    typedef complex<ld> point;
    typedef pair<int, int> pii;
    typedef pair<pii, int> piii;
    typedef vector<int> vi;
    
    #define CLR(x,y) memset(x,y,sizeof(x))
    #define CPY(x,y) memcpy(x,y,sizeof(x))
    #define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))
    #define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))
    #define debug(a) cout << #a" = " << (a) << endl;
    #define debugarry(a, n) for (int i = 0; i < (n); i++) { cout << #a"[" << i << "] = " << (a)[i] << endl; }
    
    #define mp(x,y) make_pair(x,y)
    #define pb(x) push_back(x)
    #define lowbit(x) (x&(-x))
    
    #define MID(x,y) (x+((y-x)>>1))
    #define ls (idx<<1)
    #define rs (idx<<1|1)
    #define lson ls,l,mid
    #define rson rs,mid+1,r
    
    template<class T>
    inline bool read(T &n)
    {
        T x = 0, tmp = 1;
        char c = getchar();
        while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
        if(c == EOF) return false;
        if(c == '-') c = getchar(), tmp = -1;
        while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
        n = x*tmp;
        return true;
    }
    template <class T>
    inline void write(T n)
    {
        if(n < 0)
        {
            putchar('-');
            n = -n;
        }
        int len = 0,data[20];
        while(n)
        {
            data[len++] = n%10;
            n /= 10;
        }
        if(!len) data[len++] = 0;
        while(len--) putchar(data[len]+48);
    }
    //-----------------------------------
    int n;
    vector<int> ans;
    
    bool judge(int x,int y)
    {
        string str;
        cout<<1<<" "<<x<<" "<<y<<endl;
        cout.flush();
        cin>>str;
        if(str[0]=='Y') return 1;
        return 0;
    }
    
    int main()
    {
        bool flag = true;
        int l, r, mid, pos, cnt;
        scanf("%d",&n);
        ans.push_back(1);
        for (int i = 2; i <= n; i++)
        {
            cnt = 0;
            l = 0; r = i-2; pos = 0;
            while (l <= r)
            {
                mid = (l+r)>>1; cnt++;
                if (cnt > 100000)
                {
                    flag = false;
                    break;
                }
                if (judge(ans[mid], i))
                {
                    l = mid+1;
                    pos = mid+1;
                }
                else r = mid-1;
            }
            if (!flag) break;
            ans.insert(ans.begin()+pos, i);
        }
        printf("0 ");
        if (flag)
        {
            for (int i = 0; i <= n-2; i++) printf("%d ", ans[i]);
            printf("%d\n", ans[n-1]);
        }
        else
        {
            for (int i = 0; i <= n-2; i++) printf("0 ");
            printf("0\n");
        }
        //system("pause");
        return 0;
    }