UVALive 6955 Finding Lines（随机化优化）题解

Posted 2021-01-01 kirinsb

题意：问你是否有一条直线满足这条直线上的点个数与总个数之比不小于p

思路：解法太暴力了，直接随机取两个数，如果能满足条件就输出可以，否则继续随机直到达到300次。证明一下为什么可以随机化，题目给出可能有P >=20的点在线上，假设最惨的情况P = 20，有100个点，所以我们选一次选不到这条直线的概率为 （80 * 79）/（100 * 99），简单点我们记做0.64，那么两次我们找不到这条线的概率为0.64*.0.64……所以我们随机300次，这样选不到这条线的概率就很小很小了，除非RP不行啊。

代码：

#include<ctime>
#include<cmath>
#include<cstdio>
#include<iostream>
#include<algorithm>
#define ll long long
#define ull unsigned long long
using namespace std;
const int maxn = 100000 + 10;
const int seed = 131;
const int MOD = 100013;
const int INF = 0x3f3f3f3f;
struct node{
    int x, y;
}p[maxn];
bool judge(int u, int v,int i){
    int x1 = p[u].x - p[i].x, x2 = p[v].x - p[i].x;
    int y1 = p[u].y - p[i].y, y2 = p[v].y - p[i].y;
    return y1 * x2 == y2 * x1;
}
int main(){
    int n, per;
    while(~scanf("%d%d", &n, &per)){
        per = (int)ceil((double)per / 100 * n);
        for(int i = 0; i < n; i++){
            scanf("%d%d", &p[i].x, &p[i].y);
        }
        if(n <= 2){
            printf("possible
");
            continue;
        }
        srand(time(0));
        int Time = 0;
        bool flag = false;
        while(Time <= 300){
            int cnt = 2;
            int u = rand() % n;
            int v = rand() % n;
            while(v == u) v = rand() % n;
            for(int i = 0; i < n; i++){
                if(i == v || i == u) continue;
                if(judge(u, v, i)) cnt++;
                if(cnt >= per){
                    flag = true;
                    break;
                }
            }
            if(flag) break;
            Time++;
        }
        if(flag)
            printf("possible
");
        else
            printf("impossible
");
    }
    return 0;
}