HDU 5762 [简单复杂度分析]

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Description

给定n个点，求是否存在两对点曼哈顿距离相等。
0≤xi,yi≤105

Solution

曼哈顿距离的个数应该也是O(n)的啦。。。

#include <bits/stdc++.h>using namespace std;const int N = 201010;inline char get(void) {    static char buf[100000], *S = buf, *T = buf;    if (S == T) {        T = (S = buf) + fread(buf, 1, 100000, stdin);        if (S == T) return EOF;    }    return *S++;}inline void read(int &x) {    static char c; x = 0;    for (c = get(); c < '0' || c > '9'; c = get());    for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';}int test, n, m, d, ans;struct Point {    int x, y;    Point (int _x = 0, int _y = 0):x(_x), y(_y) {}};Point a[N];int vis[N];inline int Dis(Point &a, Point &b) {    return fabs(a.x - b.x) + fabs(a.y - b.y);}int main(void) {    freopen("1.in", "r", stdin);    read(test);    for (int t = 1; t <= test; t++) {        read(n); read(m); ans = 0;        for (int i = 1; i <= n; i++) {            read(a[i].x); read(a[i].y);        }        for (int i = 1; i <= n; i++) {            for (int j = i + 1; j <= n; j++) {                d = Dis(a[i], a[j]);                if (vis[d] == t) {                    ans = 1; break;                }                vis[d] = t;            }            if (ans) break;        }        puts(ans ? "YES" : "NO");    }}

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