【POJ 1755】n*n的半平面交算法 适用于不等式交

Triathlon

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 5519 Accepted: 1405

Description

Triathlon is an athletic contest consisting of three consecutive sections that should be completed as fast as possible as a whole. The first section is swimming, the second section is riding bicycle and the third one is running. 

The speed of each contestant in all three sections is known. The judge can choose the length of each section arbitrarily provided that no section has zero length. As a result sometimes she could choose their lengths in such a way that some particular contestant would win the competition. 

Input

The first line of the input file contains integer number N (1 <= N <= 100), denoting the number of contestants. Then N lines follow, each line contains three integers Vi, Ui and Wi (1 <= Vi, Ui, Wi <= 10000), separated by spaces, denoting the speed of ith contestant in each section.

Output

For every contestant write to the output file one line, that contains word "Yes" if the judge could choose the lengths of the sections in such a way that this particular contestant would win (i.e. she is the only one who would come first), or word "No" if this is impossible.

Sample Input

910 2 610 7 35 6 73 2 76 2 63 5 78 4 610 4 21 8 7

Sample Output

YesYesYesNoNoNoYesNoYes

Source

Northeastern Europe 2000

#define DeBUG#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <string>#include <set>#include <sstream>#include <map>#include <list>#include <bitset>using namespace std ;#define zero {0}#define INF 0x3f3f3f3f#define EPS 1e-10#define TRUE true#define FALSE falsetypedef long long LL;const double PI = acos(-1.0);//#pragma comment(linker, "/STACK:102400000,102400000")inline int sgn(double x){    return fabs(x) < EPS ? 0 : (x < 0 ? -1 : 1);}#define N 100005// Point classstruct Point;typedef Point Vec;struct Point{    double x, y;    Point () {}    Point(double a, double b)    {        x = a;        y = b;    }};void initial(Point p[], int &pCnt) //联机算法,初始化无限大平面,切割多边形从1开始{    p[1] = Point(0, 0);    p[2] = Point(0, 1000000);    p[3] = Point(1000000, 1000000);    p[4] = Point(1000000, 0);    p[5] = p[1];    p[0] = p[4];    pCnt = 4;}Point intersect(Point x, Point y, double a, double b, double c){    double u = fabs(a * x.x + b * x.y + c);    double v = fabs(a * y.x + b * y.y + c);    return Point( (x.x * v + y.x * u) / (u + v) , (x.y * v + y.y * u) / (u + v) );}int cut(Point p[], int &pCnt, double a, double b , double c) //p为多边形数组{    Point q[1000];//队列最大长度    int  curCnt = 0;    for (int i = 1; i <= pCnt; ++i)    {        if (sgn(a * p[i].x + b * p[i].y + c) > 0)q[++curCnt] = p[i];        //注意这里一定是大于0，而不是大于等于0,根据题意变化,下面的本来就没有=        else        {            if (sgn(a * p[i - 1].x + b * p[i - 1].y + c) > 0)            {                q[++curCnt] = intersect(p[i], p[i - 1], a, b, c);            }            if (sgn(a * p[i + 1].x + b * p[i + 1].y + c) > 0)            {                q[++curCnt] = intersect(p[i], p[i + 1], a, b, c);            }        }    }    for (int i = 1; i <= curCnt; ++i)        p[i] = q[i];    p[curCnt + 1] = q[1];    p[0] = p[curCnt];    pCnt = curCnt;    return curCnt;}struct Node{    double x, y, z;};Node person[200];int main(){#ifdef DeBUGs    freopen("C:\\Users\\Sky\\Desktop\\1.in", "r", stdin);#endif    int n;    while (scanf("%d", &n) + 1)    {        for (int i = 0; i < n; i++)        {            scanf("%lf%lf%lf", &person[i].x, &person[i].y, &person[i].z);        }        Point p[200];        int num;        double a, b, c;        for (int i = 0; i < n; i++)        {            bool flag = true;            initial(p,num);            for (int j = 0; j < n; j++)            {                if (i == j)                    continue;                c = 1 / person[j].z - 1 / person[i].z;                a = 1 / person[j].x - 1 / person[i].x;                b = 1 / person[j].y - 1 / person[i].y;                if (cut(p, num, a, b, c) <= 2)                {                    flag = false;                    break;                }            }            if (flag)                printf("Yes\n");            else                printf("No\n");        }    }    return 0;}