Codeforces Round #504 E - Down or Right 交互题

1023E

题意：

　　交互题。在一个有障碍地图中，问如何走才能从（1，1）走到（n，n),只能向右或者向左走。每次询问两个点，回复你这两个点能不能走通。

思路：

　　只用最多2*n-2次询问。从（1，1），能向右走就向右走，不能就向下走，直到走到斜对角线上。从（n，n）出发，能向上走就向上走，不能就向左走，直到走到斜对角线上。

　　因为保证有路，所以最后输出（1，1）出发的正向路径，加上从（n，n）出发的反向路径。

#include <algorithm>

#include  <iterator>

#include  <iostream>

#include   <cstring>

#include   <cstdlib>

#include   <iomanip>

#include    <bitset>

#include    <cctype>

#include    <cstdio>

#include    <string>

#include    <vector>

#include     <cmath>

#include     <queue>

#include      <list>

#include       <map>

#include       <set>

using namespace std;

//#pragma GCC optimize(3)

//#pragma comment(linker, "/STACK:102400000,102400000")  //c++

#define lson (l , mid , rt << 1)

#define rson (mid + 1 , r , rt << 1 | 1)

#define debug(x) cerr << #x << " = " << x << "\n";

#define pb push_back

#define pq priority_queue

typedef long long ll;

typedef unsigned long long ull;

typedef pair<ll ,ll > pll;

typedef pair<int ,int > pii;

typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q

//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q

#define fi first

#define se second

//#define endl '\n'

#define OKC ios::sync_with_stdio(false);cin.tie(0)

#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行

#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)

//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //

const ll nmos = 0x80000000;  //-2147483648

const int inf = 0x3f3f3f3f;

const ll inff = 0x3f3f3f3f3f3f3f3f; //

const int mod = 1e9+;

const double PI=acos(-1.0);

template<typename T>

inline T read(T&x){

    x=;int f=;char ch=getchar();

    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();

    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();

    return x=f?-x:x;

}

// #define _DEBUG;         //*//

#ifdef _DEBUG

freopen("input", "r", stdin);

// freopen("output.txt", "w", stdout);

#endif

/*-----------------------showtime----------------------*/

            const int maxn = ;

            int vis[maxn];

            int n;

            int get(int x,int y,int flag){

                char g[];

                if(flag)

                    cout<<"? "<<x<<" "<<y<<" "<<n<<" "<<n<<endl;

                else cout<<"? "<<<<" "<<<<" "<<x<<" "<<y<<endl;

                cin>>g;

                if(g[] == 'Y')return ;

                else if(g[] == 'N') return ;

                return ;

            }

            vector<int>up,down;

            void solve(){

                int x = , y = ;

                while(x+y<=n+){

                    if(get(x,y,)){

                        y++;up.pb();

                    }

                    else {

                        x++;up.pb();

                    }

                }

                x = n-,y = n;

                while(x+y>n){

                    if(get(x,y,)){

                        x--;down.pb();

                    }

                    else {

                        y--;down.pb();

                    }

                }

            }

int main(){

            cin>>n;

            solve();

            string ans = "";

            for(int i=; i<up.size(); i++){

                int tmp = up[i];

                if(tmp==){

                    ans+="D";

                }

                else ans += "R";

            }

            for(int i=down.size() - ; i>=; i--){

                int tmp = down[i];

                if(tmp==){

                    ans+="D";

                }

                else ans += "R";

            }

            cout<<"! "<<ans<<endl;

            return ;

}

1023E