Codeforces Round #371 (Div. 2)D. Searching Rectangles交互题
2016-09-20 14:48
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题目大意:Filya在边长为n的正方形里面, 画了两个矩形。这两个矩形的边与正方形的边平行,且两个长方形没有不能相交(边可以重合)。但是Filya忘记他把两个矩形画在正方形的哪边了,所以他问Sonya在以(x1, y1)为左下顶点和以(x2, y2)为右上顶点的矩形里面,完整包含了几个他之前画的矩形, 问的次数不能超过200次。让你来模拟这个过程。你的输入首先是n(正方形的边长),然后由计算机来问你,你的输入就是回答的答案。
方法二分即可,但是搞不懂为什么用cin, cout就不用清空缓冲区,这个题目清空缓冲区的意义何在?
方法二分即可,但是搞不懂为什么用cin, cout就不用清空缓冲区,这个题目清空缓冲区的意义何在?
#include<bits/stdc++.h> using namespace std; struct abc { int x1,y1,x2,y2; void print(){ cout<<x1<<" "<<y1<<" "<<x2<<" "<<y2<<" "; } }; int query(int x1,int y1,int x2,int y2) { if(x1>x2)swap(x1,x2); if(y1>y2)swap(y1,y2); cout<<"? "<<x1<<" "<<y1<<" "<<x2<<" "<<y2<<endl; int ans;cin>>ans; return ans; } abc fi(int xx1,int yy1,int xx2,int yy2) { abc Ans; int l=yy1,r=yy2,ans=yy1; while(l<=r) { int mid=(l+r)/2; if(query(xx1,yy1,xx2,mid)<1) l=mid+1; else r=mid-1,ans=mid; } Ans.y2=ans; l=yy1,r=yy2,ans=yy1; while(l<=r) { int mid=(l+r)/2; if(query(xx1,mid,xx2,yy2)<1) r=mid-1; else l=mid+1,ans=mid; } Ans.y1=ans; l=xx1,r=xx2,ans=xx1; while(l<=r) { int mid=(l+r)/2; if(query(xx1,yy1,mid,yy2)<1) l=mid+1; else r=mid-1,ans=mid; } Ans.x2=ans; l=xx1,r=xx2,ans=xx1; while(l<=r) { int mid=(l+r)/2; if(query(mid,yy1,xx2,yy2)<1) r=mid-1; else l=mid+1,ans=mid; } Ans.x1=ans; return Ans; } int main() { int n; scanf("%d",&n); int l=1,r=n,ans=1; while(l<=r) { int mid=(l+r)/2; if(query(1,1,mid,n)<1) l=mid+1; else r=mid-1,ans=mid; } if(query(1,1,ans,n)==1&&query(ans+1,1,n,n)==1) { abc r1=fi(1,1,ans,n); abc r2=fi(ans+1,1,n,n); cout<<"! "; r1.print(); r2.print(); return 0; } l=1,r=n,ans=1; while(l<=r) { int mid=(l+r)/2; if(query(1,1,n,mid)<1) l=mid+1; else r=mid-1,ans=mid; } abc r1 = fi(1,1,n,ans); abc r2 = fi(1,ans+1,n,n); cout<<"! "; r1.print(); r2.print(); }
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