HDU1432+几何

题意：给N个点，求最多有多少个点在同一直线上

方法一：求出所有能形成的直线，两两比较，统计最多有多少条重合。

  1 #include<stdio.h>
  2 #include<stdlib.h>
  3 #include<string.h>
  4 #include<algorithm>
  5 #include<set>
  6 #include<iostream>
  7 using namespace std;
  8 typedef long long int64;
  9 const int64 maxn = 701;
 10 const int64 maxm = 490005;
 11 const int64 inf = -123456789;
 12 struct Point64{
 13     int x,y;
 14 }pnt[ maxn ];
 15 struct Node{
 16     int64 a,b,c;
 17     int x,y;
 18 }tp[ maxm ];
 19 
 20 bool s[ maxn ];
 21 //set<int>s;
 22 
 23 void init( int n ){
 24     for( int i=0;i<n;i++ )
 25         s[ i ] = false;
 26 }
 27 
 28 int64 cmp( Node p1,Node p2 ){
 29     if( p1.a!=p2.a ) return p1.a<p2.a;
 30     else if( p1.b!=p2.b ) return p1.b<p2.b;
 31     else return p1.c<p2.c;
 32 }
 33 
 34 int64 gcd( int64 a,int64 b ){
 35     int64 r;
 36     while( b ){
 37         r = a%b;
 38         a = b;
 39         b = r;
 40     }
 41     return a;
 42 }
 43 
 44 void solve_gcd( int id ){
 45     int64 Gcd = gcd( tp[ id ].a,tp[ id ].b );
 46     Gcd = gcd( tp[ id ].c,Gcd );
 47     tp[ id ].a /= Gcd;
 48     tp[ id ].b /= Gcd;
 49     tp[ id ].c /= Gcd;
 50 }    
 51 
 52 int main(){
 53     int n;
 54     while( scanf("%d",&n)!=EOF ){
 55         for( int64 i=0;i<n;i++ ){
 56             cin>>pnt[i].x>>pnt[i].y;
 57             //scanf("%lld%lld",&pnt[i].x,&pnt[i].y);
 58             s[ i ] = false;
 59         }
 60     if( n==1||n==2 ){
 61         printf("%d\n",n);
 62         continue;
 63     } 
 64         int cnt = 0;
 65         for( int i=0;i<n;i++ ){
 66             for( int j=i+1;j<n;j++ ){
 67                 tp[ cnt ].a = pnt[ i ].y - pnt[ j ].y;
 68                 tp[ cnt ].b = pnt[ j ].x - pnt[ i ].x;
 69                 tp[ cnt ].c = pnt[ i ].x*pnt[ j ].y - pnt[ j ].x*pnt[ i ].y;
 70                 tp[ cnt ].x = i;
 71                 tp[ cnt ].y = j;
 72                 solve_gcd( cnt );
 73                 cnt ++;
 74             }
 75         }
 76         sort( tp,tp+cnt,cmp );
 77         int64 ans = 2;
 78         int64 ans_tp = 0;
 79         int64 prea = inf;
 80         int64 preb = inf;
 81         int64 prec = inf;
 82         for( int i=0;i<cnt;i++ ){
 83             if(( prea!=tp[ i ].a || preb!=tp[ i ].b || prec!=tp[ i ].c )){
 84                 ans_tp = 2;
 85                 prea = tp[ i ].a;
 86                 preb = tp[ i ].b;
 87                 prec = tp[ i ].c;
 88                 memset( s,false,sizeof( s ) );
 89                 s[ tp[i].x ] = true;
 90                 s[ tp[i].y ] = true;
 91             }
 92             else {
 93                 if( s[ tp[i].x ]==false ) {
 94                     ans_tp ++;
 95                     s[ tp[i].x ] = true;
 96                 }
 97                 if( s[ tp[i].y ]==false ) {
 98                     ans_tp ++;
 99                     s[ tp[i].y ] = true;
100                 }
101                 ans = max( ans,ans_tp );
102             }
103         }
104         cout<<ans<<endl;
105         //printf("%lld\n",ans);
106     }
107     return 0;
108 }

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