HDU 5762 Teacher Bo （暴力）

Teacher Bo

题目链接：

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

Source

2016 Multi-University Training Contest 3


##题意： 给出平面上的N个点，求是否一个四元组(A,B,C,D)满足A\B之间的曼哈顿距离等于C\D之间距离. 条件(A
##题解： 这个题让我懵比的两个小时... 一开始先把曼哈顿距离想成了abs(dx+dy)，然后可以只考虑每个点的坐标和，从而误把题目转化为求一个数组中是否有两对数的和相等. 由于规模是1e5,然后一直在想O(nlogn)的算法...
首先，曼哈顿距离的正确定义是abs(dx)+abs(dy). 其次，题目的数据规模确实不能跑O(n^2)的算法，但是题目限制了坐标的范围为[0,1e5], 这意味着任意两点的曼哈顿距离只可能有2e5种可能，暴力最多只会判断2e5个组合，所以不会超时. 所以直接暴力判断所有组合即可，当出现重复的距离时停止判断(先离线数据).
这个题懵比这么久真的是非常不应该，所以以后卡题的时候，一定要重新读题和尝试转换思考角度.
##代码： ``` cpp #include #include #include #include #include #include #include #include #include #define LL long long #define eps 1e-8 #define maxn 251000 #define mod 100000007 #define inf 0x3f3f3f3f #define IN freopen("in.txt","r",stdin); using namespace std;

int n,m;
struct point {
int x,y;
}p[maxn];
bool dis[maxn];

int get_ans(point a, point b) {
return abs(a.x-b.x) + abs(a.y-b.y);
}

int main(int argc, char const *argv[])
{
//IN;

int t; cin >> t;
while(t--)
{
    memset(dis, 0, sizeof(dis));
    cin >> n >> m;

    int flag = 0;

    for(int i=1; i<=n; i++) {
        int x,y; scanf("%d %d", &x,&y);
        p[i].x=x; p[i].y=y;
    }

    for(int i=1; i<=n; i++) {
        for(int j=1; j<i; j++) {
            int tmp = get_ans(p[i],p[j]);
            if(dis[tmp]) {
                flag = 1;
                break;
            }
            dis[tmp] = 1;
        }
        if(flag) break;
    }

    if(flag) puts("YES");
    else puts("NO");
}

return 0;

}
posted @ 2016-07-26 22:14  Sunshine_tcf  阅读(342)  评论(0)  编辑  收藏  举报