CF 714D Searching Rectangles 交互题 二分

Posted 2020-10-02

题意:交互题,已知一个n*n的地图,电脑藏了两个不相交的矩形(矩形的边和坐标轴平行).
每次向电脑询问左下[x1,y1]右上[x2,y2]的矩形内完全包含多少个藏着的矩形.
n<=2^16.请在200次内猜出两个矩形左下,右上坐标.

先二分出两个不相交矩形的分界线,要么横着要么竖着.
接在在每一快中二分出矩形的四条边即可(最左,右,上,下边界)

#include <bits/stdc++.h>
using namespace std;
 
int n,ans[20],cnt=0;
inline int out(int x,int y,int fx,int fy)
{
    if(x>fx||y>fy)return 0;
    printf("? %d %d %d %d\n",x,y,fx,fy);
    fflush(stdout);
    int num;
    cin>>num;
    return num;
}
inline void solve(int x,int y,int fx,int fy)
{
    int l=x,r=fx,Ans=x;
       while(l<=r)
    {
        int mid=(l+r)>>1;
        int f=out(mid,y,fx,fy);
        if(f)
            l=mid+1,Ans=mid;
        else
            r=mid-1;
    }
    ans[++cnt]=Ans;
    l=y,r=fy,Ans=y;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        int f=out(x,mid,fx,fy);
        if(f==0)r=mid-1;
        else l=mid+1,Ans=mid;
    }
    ans[++cnt]=Ans;
    l=x,r=fx,Ans=fx;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        int f=out(x,y,mid,fy);
        if(f==0)l=mid+1;
        else r=mid-1,Ans=mid;
    }
    ans[++cnt]=Ans;
    l=y,r=fy,Ans=fy;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        int f=out(x,y,fx,mid);
        if(f==0)l=mid+1;
        else r=mid-1,Ans=mid;
    }
    ans[++cnt]=Ans;
}
int main()
{
    cin>>n;
    int l=1,r=n,flag=0;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        int f=out(1,1,n,mid),g=out(1,mid+1,n,n);
        if(f&&g)flag=1;
        if(f==0)l=mid+1;
        else r=mid-1;
    }
    if(flag)
    {
        solve(1,1,n,l);
        solve(1,l+1,n,n);
    }
    else
    {
        l=1,r=n;
        while(l<=r)
        {
            int mid=(l+r)>>1;
            int f=out(1,1,mid,n);
            if(f==0)l=mid+1;
            else r=mid-1;
        }
        solve(1,1,l,n);
        solve(l+1,1,n,n);
    }
    printf("! ");
    for(int i=1;i<=8;i++)printf("%d ",ans[i]);
        printf("\n");
    return 0;
}