随机算法 - HNU 13348 Finding Lines

/*
* 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;
}
/*

*/

// 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;
}

#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;
}