• hdu 5762 Teacher Bo 暴力


    Teacher Bo

    题目连接:

    http://acm.hdu.edu.cn/showproblem.php?pid=5762

    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".
    

    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).

    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

    Hint

    题意

    让你找到两组不同的点,使得他们的曼哈顿距离相同

    题解:

    考虑一种暴力,每次枚举两两点对之间的曼哈顿距离,并开一个桶记录每种距离是否出现过,如果某次枚举出现了以前出现的距离就输 YES ,否则就输 NO .

    注意到曼哈顿距离只有 O(M) 种,根据鸽笼原理,上面的算法在 O(M)步之内一定会停止.所以是可以过得.

    一组数据的时间复杂度 O(min{N^2,M})
    ​​ ,M}) .

    代码

    #include <bits/stdc++.h>
    
    using namespace std;
    
    const int N=100010;
    int x[N],y[N],vis[N*4];
    
    int main()
    {
        int T;
        scanf("%d",&T);
        while(T--)
        {
            int n,m;
            scanf("%d%d",&n,&m);
            int flag=1,mx=m*4+10,cnt=0;
            for(int i=0;i<=mx;i++) vis[i]=0;
            for(int i=0;i<n;i++)
                scanf("%d%d",&x[i],&y[i]);
            for(int i=0;i<n;i++)
            {
                for(int j=i+1;j<n;j++)
                {
                    if(cnt>mx)
                    {
                        flag=0;
                        break;
                    }
                    if(vis[abs(x[i]-x[j])+abs(y[i]-y[j])])
                    {
                        flag=-1;
                        break;
                    }
                    vis[abs(x[i]-x[j])+abs(y[i]-y[j])]=1;cnt++;
                }
                if(flag<1) break;
            }
            if(flag==-1) printf("YES
    ");else printf("NO
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/5708491.html
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