POJ 1755 Triathlon 半平面交
题意：铁人三项比赛，给出ｎ个人进行每一项的速度vi, ui, wi; 对每个人判断，通过改变3项比赛的路程，是否能让该人获胜（严格获胜）。思路：题目实际上是给出了n个式子方程，Ti = Ai * x + Bi * y + Ci * z ， 0 < i < n
要判断第i个人能否获胜，即判断不等式组 Tj - Ti > 0, 0 < j < n && j != i 有解
即 （Aj - Ai）* x + （Bj - Bi） * y + ( Cj - Ci ) * z > 0, 0 < j < n && j != i 有解
由于 z > 0, 所以 可以两边同时除以 z， 将 x / z, y / z 分别看成 x和 y ， 这样就化三维为二维，可用半平面交判断是否存在解了，
对每个人构造一次，求一次半平面交即可。
关键是根据这个斜率式子怎么搞成向量的。需要想一想。
然后注意的是半平面交出来是单独一个点是不行的。
因为题目要求的是严格胜出
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <map>
#include <sstream>
#include <queue>
#include <vector>
#define MAXN 111111
#define MAXM 211111
#define PI acos(-1.0)
#define eps 1e-8
#define INF 1e10
using namespace std;
int dblcmp(double d)
{
if (fabs(d) < eps) return 0;
return d > eps ? 1 : -1;
}
struct point
{
double x, y;
point(){}
point(double _x, double _y):
x(_x), y(_y){};
void input()
{
scanf("%lf%lf",&x, &y);
}
double dot(point p)
{
return x * p.x + y * p.y;
}
double distance(point p)
{
return hypot(x - p.x, y - p.y);
}
point sub(point p)
{
return point(x - p.x, y - p.y);
}
double det(point p)
{
return x * p.y - y * p.x;
}
bool operator == (point a)const
{
return dblcmp(a.x - x) == 0 && dblcmp(a.y - y) == 0;
}
bool operator < (point a)const
{
return dblcmp(a.x - x) == 0 ? dblcmp(y - a.y) < 0 : x < a.x;
}
}p[MAXN];
struct line
{
point a,b;
line(){}
line(point _a,point _b)
{
a=_a;
b=_b;
}
bool parallel(line v)
{
return dblcmp(b.sub(a).det(v.b.sub(v.a))) == 0;
}
point crosspoint(line v)
{
double a1 = v.b.sub(v.a).det(a.sub(v.a));
double a2 = v.b.sub(v.a).det(b.sub(v.a));
return point((a.x * a2 - b.x * a1) / (a2 - a1), (a.y * a2 - b.y * a1) / (a2 - a1));
}
bool operator == (line v)const
{
return (a == v.a) && (b == v.b);
}
};
struct halfplane:public line
{
double angle;
halfplane(){}
//表示向量 a->b逆时针(左侧)的半平面
halfplane(point _a, point _b)
{
a = _a;
b = _b;
}
halfplane(line v)
{
a = v.a;
b = v.b;
}
void calcangle()
{
angle = atan2(b.y - a.y, b.x - a.x);
}
bool operator <(const halfplane &b)const
{
return dblcmp(angle - b.angle) < 0;
}
};
struct polygon
{
int n;
point p[MAXN];
line l[MAXN];
double area;
void getline()
{
for (int i = 0; i < n; i++)
{
l[i] = line(p[i], p[(i + 1) % n]);
}
}
void getarea()
{
area = 0;
int a = 1, b = 2;
while(b <= n - 1)
{
area += p[a].sub(p[0]).det(p[b].sub(p[0]));
a++;
b++;
}
area = fabs(area) / 2;
}
}convex;
bool judge(point a, point b, point o)
{
return dblcmp(a.sub(o).det(b.sub(o))) <= 0; //此处有等于号代表的是求出的半平面交为一个点不合法，去掉等于号则代表交成一个点也行
}
struct halfplanes
{
int n;
halfplane hp[MAXN];
point p[MAXN];
int que[MAXN];
int st, ed;
void push(halfplane tmp)
{
hp[n++] = tmp;
}
void unique()
{
int m = 1, i;
for (i = 1; i < n;i++)
{
if (dblcmp(hp[i].angle - hp[i - 1].angle))hp[m++] = hp[i];
else if (dblcmp(hp[m - 1].b.sub(hp[m - 1].a).det(hp[i].a.sub(hp[m - 1].a)) > 0))hp[m - 1] = hp[i];
}
n = m;
}
bool halfplaneinsert()
{
int i;
for (i = 0; i < n; i++) hp[i].calcangle();
sort(hp, hp + n);
unique();
que[st = 0] = 0;
que[ed = 1] = 1;
p[1] = hp[0].crosspoint(hp[1]);
for (i = 2; i < n; i++)
{
while (st < ed && judge(hp[i].b, p[ed], hp[i].a)) ed--;
while (st < ed && judge(hp[i].b, p[st + 1], hp[i].a)) st++;
que[++ed] = i;
if (hp[i].parallel(hp[que[ed - 1]])) return false;
p[ed] = hp[i].crosspoint(hp[que[ed - 1]]);
}
while (st < ed && judge(hp[que[st]].b, p[ed], hp[que[st]].a)) ed--;
while (st < ed && judge(hp[que[ed]].b, p[st + 1], hp[que[ed]].a)) st++;
if (st + 1 >= ed)return false;
return true;
}
void getconvex(polygon &con)
{
p[st] = hp[que[st]].crosspoint(hp[que[ed]]);
con.n = ed - st + 1;
int j = st, i = 0;
for (; j <= ed; i++, j++)
{
con.p[i] = p[j];
}
}
}h;
int A[MAXN], B[MAXN], C[MAXN];
int n;
int main()
{
double xa, xb, ya, yb;
scanf("%d", &n);
for(int i = 0; i < n; i++) scanf("%d%d%d", &A[i], &B[i], &C[i]);
for(int i = 0; i < n; i++)
{
int flag = 0;
h.n = 0;
h.push(halfplane(point(0, 0), point(INF, 0)));
h.push(halfplane(point(INF, 0), point(INF, INF)));
h.push(halfplane(point(INF, INF), point(0, INF)));
h.push(halfplane(point(0, INF), point(0, 0)));
for(int j = 0; j < n; j++)
{
if(j == i) continue;
double a = 1.0 / A[j] - 1.0 / A[i];
double b = 1.0 / B[j] - 1.0 / B[i];
double c = 1.0 / C[j] - 1.0 / C[i];
int d1 = dblcmp(a);
int d2 = dblcmp(b);
int d3 = dblcmp(c);
if(!d1)
{
if(!d2)
{
if(d3 <= 0)
{
flag = 1;
break;
}
continue;
}
xa = 0, xb = d2;
ya = yb = -c / b;
}
else
{
if(!d2)
{
xa = xb = -c / a;
ya = 0, yb = -d1;
}
else
{
xa = 0;
ya = -c / b;
xb = d2;
yb = -(c + a * xb) / b;
}
}
h.push(halfplane(point(xa, ya), point(xb, yb)));
}
if(flag || !h.halfplaneinsert() ) puts("No");
else puts("Yes");
}
return 0;
}
补充：软件开发 , C++ ,
