牛客 Fruit Ninja 2018 ACM 上海大都会赛 (随机化算法)

题目链接：Fruit Ninja
比赛链接：2018 ACM 国际大学生程序设计竞赛上海大都会赛重现赛

题目描述

Fruit Ninja is a juicy action game enjoyed by millions of players around the world, with squishy,

splat and satisfying fruit carnage! Become the ultimate bringer of sweet, tasty destruction with every slash.

Fruit Ninja is a very popular game on cell phones where people can enjoy cutting the fruit by touching the screen.

In this problem, the screen is rectangular, and all the fruits can be considered as a point. A touch is a straight line cutting

thought the whole screen, all the fruits in the line will be cut.

A touch is EXCELLENT if \(\frac{M}{N} \ge x\), (\(N\) is total number of fruits in the screen, \(M\) is the number of fruits that cut by the touch, \(x\) is a real number.)
Now you are given \(N\) fruits position in the screen, you want to know if exist a EXCELLENT touch.

输入描述:

The first line of the input is \(T(1 \le T \le 100)\), which stands for the number of test cases you need to solve.
The first line of each case contains an integer \(N (1 \le N \le 10^4)\) and a real number \(x (0 < x < 1)\), as mentioned above.
The real number will have only 1 digit after the decimal point.
The next \(N\) lines, each lines contains two integers \(x_i\) and \(y_i (-10^9 ≤ x_i,y_i ≤ 10^9)\), denotes the coordinates of a fruit.

输出描述:

For each test case, output "Yes" if there are at least one EXCELLENT touch. Otherwise, output "No".

示例1

输入

  2
  5 0.6
  -1 -1
  20 1
  1 20
  5 5
  9 9
  5 0.5
  -1 -1
  20 1
  1 20
  2 5
  9 9

输出

扫描二维码关注公众号，回复： 7401231 查看本文章

  Yes
  No

题意

给定 \(N\) 个点以及一个实数 \(x\)，问是否存在一条直线，直线经过 \(N\) 个点中的 \(M\) 个点，且 \(\frac{M}{N} \ge x\)。

思路

随机化算法

随机取两个点构造直线，比较直线经过的点的个数（斜率相同）是否大于等于 \(N * x\)。

代码

#include <bits/stdc++.h>
using namespace std;
typedef double db;
const int maxn = 1e4 + 10;

class Point {
public:
    db x, y;
    Point(db x = 0, db y = 0) : x(x), y(y) {}
    void input() {
        scanf("%lf%lf", &x, &y);
    }
};
Point p[maxn];

int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        int n;
        db x;
        scanf("%d%lf", &n, &x);
        x *= n;
        for(int i = 0; i < n; ++i) {
            p[i].input();
        }
        int flag = 0;
        for(int i = 0; i < 1000; ++i) {
            int a = rand() % n, b = rand() % n;
            while(a == b) a = rand() % n;
            db cnt = 2;
            for(int j = 0; j < n; ++j) {
                if(j == a || j == b) continue;
                if((p[j].y - p[a].y) * (p[b].x - p[j].x) == (p[j].x - p[a].x) * (p[b].y - p[j].y)) {
                    ++cnt;
                }
            }
            if(cnt >= x) {
                flag = 1;
                break;
            }
        }
        if(flag) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}