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  • Codeforces Round #340 (Div. 2) D

    Description

    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:

    The variant of the polyline in the second sample:

    The variant of the polyline in the third sample:

    没什么好说的,注意情况就是

    #include<stdio.h>
    //#include<bits/stdc++.h>
    #include<string.h>
    #include<iostream>
    #include<math.h>
    #include<sstream>
    #include<set>
    #include<queue>
    #include<map>
    #include<vector>
    #include<algorithm>
    #include<limits.h>
    #define inf 0x3fffffff
    #define INF 0x3f3f3f3f
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define LL long long
    #define ULL unsigned long long
    using namespace std;
    int main()
    {
        int x1,y1,x2,y2,x3,y3;
        cin>>x1>>y1>>x2>>y2>>x3>>y3;
        if(x1==x2&&x2==x3||y1==y2&&y2==y3)
        {
            cout<<1;
        }
        else if(x1==x2&&y3<=min(y1,y2)||x1==x2&&y3>=max(y1,y2))
        {
            cout<<2;
        }
        else if(y1==y2&&x3<=min(x1,x2)||y1==y2&&x3>=max(x1,x2))
        {
            cout<<2;
        }
        else if(x1==x3&&y2<=min(y1,y3)||x1==x3&&y2>=max(y1,y3))
        {
            cout<<2;
        }
        else if(y1==y3&&x2<=min(x1,x3)||y1==y3&&x2>=max(x1,x3))
        {
            cout<<2;
        }
        else if(x2==x3&&y1<=min(y2,y3)||x2==x3&&y1>=max(y2,y3))
        {
            cout<<2;
        }
        else if(y2==y3&&x1<=min(x2,x3)||y2==y3&&x1>=max(x2,x3))
        {
            cout<<2;
        }
        else
        {
            cout<<3;
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/yinghualuowu/p/5167655.html
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