Codeforces Round #340 (Div. 2) D. Polyline (点之间关系)

D. Polyline
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

There are three points marked on the coordinate plane. The goal is to make a simple polyline, without self-intersections and self-touches, such that it passes through all these points. Also, the polyline must consist of only segments parallel to the coordinate axes. You are to find the minimum number of segments this polyline may consist of.

Input

Each of the three lines of the input contains two integers. The i-th line contains integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th point. It is guaranteed that all points are distinct.

Output

Print a single number — the minimum possible number of segments of the polyline.

Sample test(s)
Input
1 -1
1 1
1 2
Output
1
Input
-1 -1
-1 3
4 3
Output
2
Input
1 1
2 3
3 2
Output
3
Note

The variant of the polyline in the first sample: Codeforces Round #340 (Div. 2) D. Polyline (点之间关系)_第1张图片 The variant of the polyline in the second sample: Codeforces Round #340 (Div. 2) D. Polyline (点之间关系)_第2张图片 The variant of the polyline in the third sample: Codeforces Round #340 (Div. 2) D. Polyline (点之间关系)_第3张图片



题意:给你三个点,求最少使用的线段数使其链接起来


思路:暴力枚举情况即可。。



ac代码:

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
struct s
{
	int x,y;
}a[4];
int check(int q,int w,int e)
{
	if(e>min(q,w)&&e<max(q,w))
	return 0;
	return 1;
}
int main()
{
	while(scanf("%d%d%d%d%d%d",&a[1].x,&a[1].y,&a[2].x,&a[2].y,&a[3].x,&a[3].y)!=EOF)
	{
		int ans;
		if((a[1].x==a[2].x&&a[2].x==a[3].x)||(a[1].y==a[2].y&&a[2].y==a[3].y))
		ans=1;
		else if((a[1].x==a[2].x&&check(a[1].y,a[2].y,a[3].y))||(a[1].x==a[3].x&&check(a[1].y,a[3].y,a[2].y))||(a[3].x==a[2].x&&check(a[3].y,a[2].y,a[1].y)))
	    ans=2;
		else if((a[1].y==a[2].y&&check(a[1].x,a[2].x,a[3].x))||(a[1].y==a[3].y&&check(a[1].x,a[3].x,a[2].x))||(a[3].y==a[2].y&&check(a[3].x,a[2].x,a[1].x)))
		ans=2;
		else 
		ans=3;
		printf("%d\n",ans);
	}
	return 0;
}


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