poj 2318 叉积解点和直线的关系

要判断点的位置，就是利用二分法判断和分界线的相对位置，最终确定点的所在块，利用数组，判断点的位置时利用叉积算有方向的面积．

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cstdlib>//#include <cmath>#define MAX 5007using namespace std;struct Point {    int x , y;    Point ( ) : x(0) , y(0) {}    Point ( int a , int b ) : x(a),y(b){}}p;struct Line {    Point up , down;}l[MAX];int det ( Point a  , Point b ){    return a.x * b.y - a.y*b.x;}int n,m,x1,y1,x2,y2,u,d;int  id ( Point p ){    if ( det ( Point ( p.x - l[n].down.x , p.y - l[n].down.y ) ,         Point ( l[n].up.x - p.x , l[n].up.y - p.y))            > 0 ) return n+1;    int left = 1 , right = n , mid;    while ( left != right )    {        mid = left + right >> 1;       // cout << mid << endl;        //cout << det ( Point ( p.x-l[mid].down.x , p.y-l[mid].down.y ) ,           //   Point ( l[n].up.x - p.x , l[mid].up.y - p.y)) << endl;        if ( det ( Point ( p.x-l[mid].down.x , p.y-l[mid].down.y ) ,             Point ( l[mid].up.x - p.x , l[mid].up.y - p.y ))                > 0 )  left  = mid + 1;        else right = mid;    }    return left; }int main ( ){    int ans [MAX];    while ( ~scanf("%d" , &n ),n )    {        scanf ( "%d%d%d%d%d" , &m , &x1 , &y1 , &x2 , &y2 );        memset ( ans , 0 , sizeof ( ans ) );        for ( int i = 1 ; i <= n ; i++ )        {            scanf ( "%d%d" , &l[i].up.x , &l[i].down.x );            l[i].up.y = y1 , l[i].down.y =y2;        }        for ( int i = 1 ; i <= m ; i++ )        {            scanf ( "%d%d" , &p.x , &p.y );            ans[id( p )]++;        }        for ( int i = 1 ; i <= n+1 ; i++ )            printf ( "%d: %d\n" , i-1 , ans[i] );        printf ( "\n" );    }    }