随机算法 - HNU 13348 Finding Lines

 Finding Lines#

Problem's Link:  http://acm.hnu.cn/online/?action=problem&type=show&id=13348&courseid=0
#

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Mean: 

给你平面上1e5个点，让你判断是否可以找到一条直线，使得p%的点都在这条直线上。

analyse:

经典的随机算法题。

每次随机出两个点，然后对n个点进行判断，看是否有p%的点在这条直线上。

关于随机算法正确性的证明：

每次随机一个点，这个点在直线上的概率是p，p最小为20%；

随机两个点来确定一条直线，那么随机一条直线的概率就是p*p，即:4%；

也就是说，在直线上概率为0.04，不在的概率为1-0.04=0.96;

那么随机k次不在直线上的概率为0.96^k，也就是答案出现误差的概率。

这题我们随机1000次，错误的概率接近1e-18，基本是不可能事件，证毕.

Time complexity: O()

 

Source code: 

/*
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-08-02-17.06
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <ctime>
#include <bits/stdc++.h>
#define  LL long long
#define  ULL unsigned long long
using namespace std;
const LL MAXN = 100010;

LL n, p;
LL x[MAXN], y[MAXN];
int main()
{
     srand( time( NULL ) );
     while( ~scanf( "%lld %lld", &n, &p ) )
     {
           for( LL i = 0; i < n; ++i )
                 scanf( "%lld %lld", &x[i], &y[i] );
           if( n == 1 ) { puts( "possible" ); continue; }
           LL sum=0;
           if( n * p % 100 == 0 ) sum = n * p / 100;
           else sum = n * p / 100 + 1;
           LL T = 1000;
           bool flag = false;
           while( T-- )
           {
                 LL a = ( LL )rand() * ( LL )rand() % n;
                 LL b = ( LL )rand() * ( LL )rand() % n;
                 if( a == b ) continue;
                 LL num = 2;
                 for( LL j = 0; j < n; ++j )
                 {
                       if( j == a || j == b ) continue;
                       if( ( y[j] - y[a] ) * ( x[b] - x[a] ) == ( x[j] - x[a] ) * ( y[b] - y[a] ) ) num++;
                 }
                 if( num >= sum ) {flag = true; break;}
           }
           if( flag ) puts( "possible" );
           else puts( "impossible" );
     }
     return 0;
}
/*

*/

 

 另外一种据说更高效的解法：

// Time complexity: O(n/(p/100)^2)
// Memory: O(n)

/* Solution method:
*
* Select 10000 (or more) pairs of points at random, and determine the lines that go through them.
* If 20% (or more) of all points are on the same line, we expect to find it about 400 times(*).
* For all lines that we find more than 230 times (at most 43), see how many points are on it.
* Report "possible" if one of them works.
* (*) Worst case: 15 points, p=20%, exactly 3 on one line: E = 10000/35 = 285.7, sigma = 16.7
*/

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <map>
using namespace std;

const int N = 1000000;
const int samples = 10000, threshold = 230;

int X[N], Y[N];

int gcd (int a, int b)
{    return b ? gcd(b, a%b) : a;
}

struct line
{    long long a, b, c; // ax+by=c
   line() {}
   // Construct line through (x1,y1) and (x2,y2)
   line (int x1, int y1, int x2, int y2)
   {    int d = gcd(abs(x1-x2), abs(y1-y2));
       if (x1-x2 < 0)
           d = -d;
       a = -(y1-y2)/d;
       b = (x1-x2)/d;
       c = a*x1 + b*y1;
   }
};

bool operator < (line L1, line L2)
{    return L1.a < L2.a || (L1.a == L2.a && (L1.b < L2.b || (L1.b == L2.b && L1.c < L2.c)));
}

map<line,int> Cnt;

// RNG (modulo 2^32)
unsigned int randnr()
{    static unsigned int rseed = 131;
   return rseed = 22695477 * rseed + 1;
}

int main()
{    int n, p, i, j, m, s;
   long long a, b, c;
   map<line,int>::iterator it;
   // Read input
   scanf("%d %d", &n, &p);
   for (i = 0; i < n; i++)
       scanf("%d %d", &X[i], &Y[i]);
   // Special case: n=1
   if (n == 1)
   {    printf("possible\n");
       return 0;
   }
   // Try a lot of random pairs of points
   for (s = 0; s < samples; s++)
   {
           i = randnr() % n;
       do
           j = randnr() % n;
       while (j == i);
       Cnt[line(X[i], Y[i], X[j], Y[j])]++; // Add to count
   }
   // Check all lines whose count is above the threshold
   for (it = Cnt.begin(); it != Cnt.end(); it++)
       if (it->second > threshold)
       {    a = (it->first).a;
           b = (it->first).b;
           c = (it->first).c;
           m = 0; // Count
           for (i = 0; i < n; i++)
               if (a*X[i] + b*Y[i] == c)
                   m++;
           if (100*m >= n*p)
           {    printf("possible\n");
               return 0;
           }
       }
   printf("impossible\n");
   return 0;
}

 再附上一种二分的做法：

#include <cstdio>
#include <cmath>
#include <set>
using namespace std;

const int N = 1000000;

int X[N], Y[N];

int gcd (int a, int b)
{    return b ? gcd(b, a%b) : a;
}

struct line
{    long long a, b, c; // ax+by=c
   line() {}
   // Construct line through (x1,y1) and (x2,y2)
   line (int x1, int y1, int x2, int y2)
   {    int d = gcd(abs(x1-x2), abs(y1-y2));
       if (x1-x2 < 0)
           d = -d;
       a = -(y1-y2)/d;
       b = (x1-x2)/d;
       c = a*x1 + b*y1;
   }
};

bool operator < (line L1, line L2)
{    return L1.a < L2.a || (L1.a == L2.a && (L1.b < L2.b || (L1.b == L2.b && L1.c < L2.c)));
}

set<line> findlines (int first, int last, int p)
{    int mid = (first+last)/2;
   int a, b, c, i, j, m;
   set<line> S, S1;
   if (p*(mid-first) <= 100) // Too few points left to split
   {    for (i = first; i < last; i++)
           for (j = i+1; j < last; j++)
               S1.insert(line(X[i], Y[i], X[j], Y[j]));
   }
   else
   {    S1 = findlines(first, mid, p);
       set<line> S2 = findlines(mid, last, p);
       S1.insert(S2.begin(), S2.end());
   }
   set<line>::iterator it;
   for (it = S1.begin(); it != S1.end(); it++)
   {    a = it->a;
       b = it->b;
       c = it->c;
       m = 0;
       for (i = first; i < last; i++)
           if (a*X[i] + b*Y[i] == c)
               m++;
       if (100*m >= p*(last-first))
           S.insert(*it);
   }
   return S;
}

int main()
{    int n, p, i;
   scanf("%d %d", &n, &p);
   for (i = 0; i < n; i++)
       scanf("%d %d", &X[i], &Y[i]);
   printf("%spossible\n", n == 1 || !findlines(0, n, p).empty() ? "" : "im");
   return 0;
}