HDU 5762 Teacher Bo(暴力+思维)

最新推荐文章于 2019-08-15 21:20:53 发布

CerberuX

最新推荐文章于 2019-08-15 21:20:53 发布

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本文探讨了一种算法问题，即在地图上找到四点，其中两点间的曼哈顿距离等于另外两点间的曼哈顿距离。文章提供了一个AC代码示例，通过计算所有可能的曼哈顿距离并检查是否存在重复的距离来解决此问题。

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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".

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

最多有2*m种曼哈顿距离，通过n计算出当前点能组成的曼哈顿距离总数，如果大于2m说明必然存在两对满足要求的点，否则暴力求解即可。

ac代码：

#include <bits/stdc++.h>
#define clr(arr,val) memset(arr,val,sizeof(arr))
#define rep(i,j,k) for(int i = j; i <= k; i++)
using namespace std;
struct point{
    int x,y;
}a[100006];
map<int, int>mp;
int main(){
    int t, m, n;
    cin  >> t;
    while(t--){
        cin >> n >> m;
        mp.clear();
        rep(i,1,n) {
            scanf("%d%d",&a[i].x, &a[i].y);
        }
        bool f = false;
        long long tmp = n*(n-1)/2;
        if(tmp > 2*m) f = true;
        else
        rep(i,1,n)
            rep(j,i+1,n){
                int l = abs(a[i].x-a[j].x)+abs(a[i].y-a[j].y);
                mp[l]++;
                if(mp[l]>1) {
                    f = true;
                    break;
                }
            }
        puts(f?"YES":"NO");
    }
    return 0;
}