AIM Tech Round 5 (rated, Div. 1 + Div. 2)C. Rectangles
AIM Tech Round 5 (rated, Div. 1 + Div. 2)C. Rectangles
C. Rectangles
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given nn rectangles on a plane with coordinates of their bottom left and upper right points. Some (n−1)(n−1) of the given nn rectangles have some common point. A point belongs to a rectangle if this point is strictly inside the rectangle or belongs to its boundary.
Find any point with integer coordinates that belongs to at least (n−1)(n−1) given rectangles.
Input
The first line contains a single integer nn (2≤n≤1326742≤n≤132674) — the number of given rectangles.
Each the next nn lines contains four integers x1x1, y1y1, x2x2 and y2y2 (−109≤x1<x2≤109−109≤x1<x2≤109, −109≤y1<y2≤109−109≤y1<y2≤109) — the coordinates of the bottom left and upper right corners of a rectangle.
Output
Print two integers xx and yy — the coordinates of any point that belongs to at least (n−1)(n−1) given rectangles.
Examples
input
Copy
3
0 0 1 1
1 1 2 2
3 0 4 1output
Copy
1 1input
Copy
3
0 0 1 1
0 1 1 2
1 0 2 1output
Copy
1 1input
Copy
4
0 0 5 5
0 0 4 4
1 1 4 4
1 1 4 4output
Copy
1 1input
Copy
5
0 0 10 8
1 2 6 7
2 3 5 6
3 4 4 5
8 1 9 2output
Copy
3 4Note
The picture below shows the rectangles in the first and second samples. The possible answers are highlighted.
The picture below shows the rectangles in the third and fourth samples.
题意：给定n个矩形，找出任意一个在n-1个矩形（或者在n个矩形）中的点
思路：维护两个数组 前缀 后缀 表示前i个矩形交的小矩形，
然后遍历一遍如果去掉某第i个矩形那么它的前缀i-1和后缀i+1这两个小矩形再取一次交，如果合法（就是两个矩形相交）就找到了
// luogu-judger-enable-o2
#include<bits/stdc++.h>
#include<unordered_set>
#define rg register ll
#define inf 2147483647
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
#define ll long long
#define maxn 200005
const double eps = 1e-6;
using namespace std;
inline ll read()
{
char ch = getchar(); ll s = 0, w = 1;
while (ch < 48 || ch>57) { if (ch == '-')w = -1; ch = getchar(); }
while (ch >= 48 && ch <= 57) { s = (s << 1) + (s << 3) + (ch ^ 48); ch = getchar(); }
return s * w;
}
inline void write(ll x)
{
if (x < 0)putchar('-'), x = -x;
if (x > 9)write(x / 10);
putchar(x % 10 + 48);
}
ll n;
struct node
{
ll x1=-inf,y1=-inf,x2=inf,y2=inf;
}a[maxn],pre[maxn],suf[maxn];
int main()
{
cin>>n;
for(rg i=1;i<=n;i++)
{
a[i].x1=read(),a[i].y1=read(),a[i].x2=read(),a[i].y2=read();
}
for(rg i=1;i<=n;i++)
{
pre[i].x1=max(pre[i-1].x1,a[i].x1);
pre[i].y1=max(pre[i-1].y1,a[i].y1);
pre[i].x2=min(pre[i-1].x2,a[i].x2);
pre[i].y2=min(pre[i-1].y2,a[i].y2);
}
for(rg i=n;i>=1;i--)
{
suf[i].x1=max(suf[i+1].x1,a[i].x1);
suf[i].y1=max(suf[i+1].y1,a[i].y1);
suf[i].x2=min(suf[i+1].x2,a[i].x2);
suf[i].y2=min(suf[i+1].y2,a[i].y2);
}
for(rg i=1;i<=n;i++)
{
ll temp1=-inf,temp2=-inf,temp3=inf,temp4=inf;
temp1=max(pre[i-1].x1,suf[i+1].x1);
temp2=max(pre[i-1].y1,suf[i+1].y1);
temp3=min(pre[i-1].x2,suf[i+1].x2);
temp4=min(pre[i-1].y2,suf[i+1].y2);
if(temp1<=temp3&&temp2<=temp4)
{
cout<<temp1<<" "<<temp2<<endl;
return 0;
}
}
return 0;
}