Problem Description
Teacher BoBo is a geography teacher in the school.One day in his class,he marked
N points in the map,the
i-th point is at
(Xi,Yi).He wonders,whether there is a tetrad
(A,B,C,D)(A<B,C<D,A≠CorB≠D) such that the manhattan distance between A and B is equal to the manhattan distance between C and D.
If there exists such tetrad,print "YES",else print "NO".
If there exists such tetrad,print "YES",else print "NO".
Input
First line, an integer
T. There are
T test cases.
(T≤50)
In each test case,the first line contains two intergers, N, M, means the number of points and the range of the coordinates. (N,M≤105).
Next N lines, the i-th line shows the coordinate of the i-th point. (Xi,Yi)(0≤Xi,Yi≤M).
In each test case,the first line contains two intergers, N, M, means the number of points and the range of the coordinates. (N,M≤105).
Next N lines, the i-th line shows the coordinate of the i-th point. (Xi,Yi)(0≤Xi,Yi≤M).
Output
T lines, each line is "YES" or "NO".
Sample Input
2 3 10 1 1 2 2 3 3 4 10 8 8 2 3 3 3 4 4
Sample Output
YES NO
估计上面也不会太认真看,我也是为了占行数。。。
题意:
搜索n个点,是否有曼哈顿距离相等的两组点。
这里如果直接暴力就是O(n^2)10^10超时,没有想到降复杂度的方法。于是我从数据范围下手。
如果有n个点,那么就有n*(n-1)条边。假设这些点都在一条直线上,(不再一条直线上曼哈顿距离相似度更高),且仅仅以相邻的点曼哈顿距离为有效距离且曼哈顿距离都不一样的,那么在m范围内,最多有i个点,i满足(i+1)*i/2<2*m。然后数据点的范围就降到m^(1/2),那么这样O(n^2)复杂度也不会超。
#include<stdio.h>
#include<iostream>
#include<string>
#include<map>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iterator>
#define N 200000
using namespace std;
struct node
{
int x,y;
bool operator<(const node&x)const{
return this->x<x.x;
}
};
map<node,int> g;
map<node,int>::iterator I;
map<node,int>::iterator J;
int f[N+10];
map<int,int> Map;
int main()
{
//freopen("in.txt","r",stdin);
int n,m;
int t;
scanf("%d",&t);
while(t--)
{
g.clear();
Map.clear();
memset(f,0,sizeof(f));
cin>>n>>m;
int fa=1;
int k=0;
int sum=0;
for(int i=0;i<n;i++)
{
int x,y;
scanf("%d%d",&x,&y);
node temp;
temp.x=x;
temp.y=y;
g[temp]++;
if(g[temp]==1)
{
f[k]=x+y;
k++;
sum++;
}
}
int temp=sqrt(m*1.0);
while(temp*temp<m)
temp++;
if(sum>temp*2)
{
printf("YES\n");
continue;
}
fa=0;
for(I=g.begin();I!=g.end();I++)
{
for(J=I;J!=g.end();J++)
{
if(J==I)
continue;
Map[abs(I->first.x-J->first.x)+abs(I->first.y-J->first.y)]++;
if(Map[abs(I->first.x-J->first.x)+abs(I->first.y-J->first.y)]>1)
fa=1;
if(fa)
break;
}
if(fa)
break;
}
if(fa)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}