Finding Lines
Problem‘s Link: http://acm.hnu.cn/online/?action=problem&type=show&id=13348&courseid=0
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;
}