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.
 
题目大意:铁人三项中,给出每个人玩铁人三项的速度,问能否通过调整这些比赛的距离(比如100米改成200米),使某个人获胜。
思路:设铁人三项的长度为x、y、z,x + y + z = 1,那么z = 1 - x - y。对于某个人cur,要满足对所有的人 i ,x / spd1[cur] + y / spd2[cur] + z / spd3[cur] < x / spd1[i] + y / spd2[i] + z / spd3[i]。
化简可得(1/spd1[i] - 1/spd1[cur] - 1/spd3[i] + 1/spd3[cur])*x + (1/spd2[i] - 1/spd2[cur] - 1/spd3[i] + 1/spd3[cur])*y + (1/spd3[i] - 1/spd3[cur]) > 0。
然后建立x+y<1和x>0,y>0,与上面的不等式,求是否存在可行域。求半平面交看有否可行域即可。
 
正在做模板,想办法把ax+by+c>0化成了两点式(见代码,要分类讨论),再做半平面交,我这种写法要EPS=1e-16才能过,丢的精度太多了。
 
代码(94MS):
 #include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std; const int MAXN = ;
const double EPS = 1e-;
const double PI = acos(-1.0);//3.14159265358979323846
const double INF = ; inline int sgn(double x) {
return (x > EPS) - (x < -EPS);
} struct Point {
double x, y, ag;
Point() {}
Point(double x, double y): x(x), y(y) {}
void read() {
scanf("%lf%lf", &x, &y);
}
bool operator == (const Point &rhs) const {
return sgn(x - rhs.x) == && sgn(y - rhs.y) == ;
}
bool operator < (const Point &rhs) const {
if(y != rhs.y) return y < rhs.y;
return x < rhs.x;
}
Point operator + (const Point &rhs) const {
return Point(x + rhs.x, y + rhs.y);
}
Point operator - (const Point &rhs) const {
return Point(x - rhs.x, y - rhs.y);
}
Point operator * (const int &b) const {
return Point(x * b, y * b);
}
Point operator / (const int &b) const {
return Point(x / b, y / b);
}
double length() {
return sqrt(x * x + y * y);
}
Point unit() {
return *this / length();
}
void print() {
printf("%.10f %.10f\n", x, y);
}
};
typedef Point Vector; double dist(const Point &a, const Point &b) {
return (a - b).length();
} double cross(const Point &a, const Point &b) {
return a.x * b.y - a.y * b.x;
}
//ret >= 0 means turn left
double cross(const Point &sp, const Point &ed, const Point &op) {
return sgn(cross(sp - op, ed - op));
} double area(const Point& a, const Point &b, const Point &c) {
return fabs(cross(a - c, b - c)) / ;
}
//counter-clockwise
Point rotate(const Point &p, double angle, const Point &o = Point(, )) {
Point t = p - o;
double x = t.x * cos(angle) - t.y * sin(angle);
double y = t.y * cos(angle) + t.x * sin(angle);
return Point(x, y) + o;
} struct Seg {
Point st, ed;
double ag;
Seg() {}
Seg(Point st, Point ed): st(st), ed(ed) {}
void read() {
st.read(); ed.read();
}
void makeAg() {
ag = atan2(ed.y - st.y, ed.x - st.x);
}
};
typedef Seg Line; //ax + by + c > 0
Line buildLine(double a, double b, double c) {
if(sgn(a) == && sgn(b) == ) return Line(Point(sgn(c) > ? - : , INF), Point(, INF));
if(sgn(a) == ) return Line(Point(sgn(b), -c/b), Point(, -c/b));
if(sgn(b) == ) return Line(Point(-c/a, ), Point(-c/a, sgn(a)));
if(b < ) return Line(Point(, -c/b), Point(, -(a + c) / b));
else return Line(Point(, -(a + c) / b), Point(, -c/b));
} void moveRight(Line &v, double r) {
double dx = v.ed.x - v.st.x, dy = v.ed.y - v.st.y;
dx = dx / dist(v.st, v.ed) * r;
dy = dy / dist(v.st, v.ed) * r;
v.st.x += dy; v.ed.x += dy;
v.st.y -= dx; v.ed.y -= dx;
} bool isOnSeg(const Seg &s, const Point &p) {
return (p == s.st || p == s.ed) ||
(((p.x - s.st.x) * (p.x - s.ed.x) < ||
(p.y - s.st.y) * (p.y - s.ed.y) < ) &&
sgn(cross(s.ed, p, s.st) == ));
} bool isIntersected(const Point &s1, const Point &e1, const Point &s2, const Point &e2) {
return (max(s1.x, e1.x) >= min(s2.x, e2.x)) &&
(max(s2.x, e2.x) >= min(s1.x, e1.x)) &&
(max(s1.y, e1.y) >= min(s2.y, e2.y)) &&
(max(s2.y, e2.y) >= min(s1.y, e1.y)) &&
(cross(s2, e1, s1) * cross(e1, e2, s1) >= ) &&
(cross(s1, e2, s2) * cross(e2, e1, s2) >= );
} bool isIntersected(const Seg &a, const Seg &b) {
return isIntersected(a.st, a.ed, b.st, b.ed);
} bool isParallel(const Seg &a, const Seg &b) {
return sgn(cross(a.ed - a.st, b.ed - b.st)) == ;
} //return Ax + By + C =0 's A, B, C
void Coefficient(const Line &L, double &A, double &B, double &C) {
A = L.ed.y - L.st.y;
B = L.st.x - L.ed.x;
C = L.ed.x * L.st.y - L.st.x * L.ed.y;
}
//point of intersection
Point operator * (const Line &a, const Line &b) {
double A1, B1, C1;
double A2, B2, C2;
Coefficient(a, A1, B1, C1);
Coefficient(b, A2, B2, C2);
Point I;
I.x = - (B2 * C1 - B1 * C2) / (A1 * B2 - A2 * B1);
I.y = (A2 * C1 - A1 * C2) / (A1 * B2 - A2 * B1);
return I;
} bool isEqual(const Line &a, const Line &b) {
double A1, B1, C1;
double A2, B2, C2;
Coefficient(a, A1, B1, C1);
Coefficient(b, A2, B2, C2);
return sgn(A1 * B2 - A2 * B1) == && sgn(A1 * C2 - A2 * C1) == && sgn(B1 * C2 - B2 * C1) == ;
} struct Poly {
int n;
Point p[MAXN];//p[n] = p[0]
void init(Point *pp, int nn) {
n = nn;
for(int i = ; i < n; ++i) p[i] = pp[i];
p[n] = p[];
}
double area() {
if(n < ) return ;
double s = p[].y * (p[n - ].x - p[].x);
for(int i = ; i < n; ++i)
s += p[i].y * (p[i - ].x - p[i + ].x);
return s / ;
}
}; void Graham_scan(Point *p, int n, int *stk, int &top) {//stk[0] = stk[top]
sort(p, p + n);
top = ;
stk[] = ; stk[] = ;
for(int i = ; i < n; ++i) {
while(top && cross(p[i], p[stk[top]], p[stk[top - ]]) >= ) --top;
stk[++top] = i;
}
int len = top;
stk[++top] = n - ;
for(int i = n - ; i >= ; --i) {
while(top != len && cross(p[i], p[stk[top]], p[stk[top - ]]) >= ) --top;
stk[++top] = i;
}
}
//use for half_planes_cross
bool cmpAg(const Line &a, const Line &b) {
if(sgn(a.ag - b.ag) == )
return sgn(cross(b.ed, a.st, b.st)) < ;
return a.ag < b.ag;
}
//clockwise, plane is on the right
bool half_planes_cross(Line *v, int vn, Poly &res, Line *deq) {
int i, n;
sort(v, v + vn, cmpAg);
for(i = n = ; i < vn; ++i) {
if(sgn(v[i].ag - v[i-].ag) == ) continue;
v[n++] = v[i];
}
int head = , tail = ;
deq[] = v[], deq[] = v[];
for(i = ; i < n; ++i) {
if(isParallel(deq[tail - ], deq[tail]) || isParallel(deq[head], deq[head + ]))
return false;
while(head < tail && sgn(cross(v[i].ed, deq[tail - ] * deq[tail], v[i].st)) > )
--tail;
while(head < tail && sgn(cross(v[i].ed, deq[head] * deq[head + ], v[i].st)) > )
++head;
deq[++tail] = v[i];
}
while(head < tail && sgn(cross(deq[head].ed, deq[tail - ] * deq[tail], deq[head].st)) > )
--tail;
while(head < tail && sgn(cross(deq[tail].ed, deq[head] * deq[head + ], deq[tail].st)) > )
++head;
if(tail <= head + ) return false;
res.n = ;
for(i = head; i < tail; ++i)
res.p[res.n++] = deq[i] * deq[i + ];
res.p[res.n++] = deq[head] * deq[tail];
res.n = unique(res.p, res.p + res.n) - res.p;
res.p[res.n] = res.p[];
return true;
} /*******************************************************************************************/ Poly poly;
Line line[MAXN], deq[MAXN];
double a[MAXN], b[MAXN], c[MAXN];
char str[];
int n, m; double calc(double x, double y) {
return (y - x) / (x * y);
} bool check(int cur) {
n = ;
line[n++] = buildLine(-, -, INF); line[n - ].makeAg();
line[n++] = buildLine(, , ); line[n - ].makeAg();
line[n++] = buildLine(, , ); line[n - ].makeAg();
for(int i = ; i < m; ++i) {
if(i == cur) continue;
line[n++] = buildLine(calc(a[i], a[cur]) + calc(c[cur], c[i]), calc(b[i], b[cur]) + calc(c[cur], c[i]), INF * calc(c[i], c[cur]));
line[n - ].makeAg();
//printf("%.10f %.10f\n", calc(a[i], a[cur]) + calc(c[cur], c[i]), calc(b[i], b[cur]) + calc(c[cur], c[i])), line[n - 1].st.print(), line[n - 1].ed.print();
}
bool flag = half_planes_cross(line, n, poly, deq);
return flag && sgn(poly.area());
} int main() {
scanf("%d", &m);
for(int i = ; i < m; ++i) scanf("%lf%lf%lf", &a[i], &b[i], &c[i]);
for(int i = ; i < m; ++i)
if(check(i)) puts("Yes");
else puts("No");
}

POJ 1755 Triathlon(线性规划の半平面交)的更多相关文章

  1. POJ 1755 Triathlon (半平面交)

    Triathlon Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4733   Accepted: 1166 Descrip ...

  2. POJ 1755 Triathlon [半平面交 线性规划]

    Triathlon Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 6912   Accepted: 1790 Descrip ...

  3. POJ 1755 Triathlon

    http://poj.org/problem?id=1755 题意:铁人三项,每个人有自己在每一段的速度,求有没有一种3条路线长度都不为0的设计使得某个人能严格获胜? 我们枚举每个人获胜,得到不等式组 ...

  4. 2018.07.03 POJ 1279Art Gallery(半平面交)

    Art Gallery Time Limit: 1000MS Memory Limit: 10000K Description The art galleries of the new and ver ...

  5. POJ 3335 Rotating Scoreboard 半平面交求核

    LINK 题意:给出一个多边形,求是否存在核. 思路:比较裸的题,要注意的是求系数和交点时的x和y坐标不要搞混...判断核的顶点数是否大于1就行了 /** @Date : 2017-07-20 19: ...

  6. POJ 1474 Video Surveillance 半平面交/多边形核是否存在

    http://poj.org/problem?id=1474 解法同POJ 1279 A一送一 缺点是还是O(n^2) ...nlogn的过几天补上... /********************* ...

  7. POJ 1279 Art Gallery 半平面交/多边形求核

    http://poj.org/problem?id=1279 顺时针给你一个多边形...求能看到所有点的面积...用半平面对所有边取交即可,模版题 这里的半平面交是O(n^2)的算法...比较逗比.. ...

  8. BZOJ1896 Equations 线性规划+半平面交+三分

    题意简述 给你\(3\)个数组\(a_i\),\(b_i\)和\(c_i\),让你维护一个数组\(x_i\),共\(m\)组询问,每次给定两个数\(s\),\(t\),使得 \[ \sum_i a_i ...

  9. POJ 1279 Art Gallery 半平面交求多边形核

    第一道半平面交,只会写N^2. 将每条边化作一个不等式,ax+by+c>0,所以要固定顺序,方便求解. 半平面交其实就是对一系列的不等式组进行求解可行解. 如果某点在直线右侧,说明那个点在区域内 ...

随机推荐

  1. HDU1398 Square Coins(生成函数)

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission( ...

  2. chromium之message_pump_win之二

    接下来分析 MessagePumpForUI上一篇分析MessagePumpWin,可以参考chromium之message_pump_win之一 根据对MessagePumpWin的分析,Messa ...

  3. Spring Cloud 微服务入门(一)--初识分布式及其发展历程

    分布式开发出现背景 当有计算机出现一段时间之后就开始有人去想如何将不同的电脑进行网络连接,而网络连接之后对于web的项目开发就探索所谓的分布式设计,同时人们也意识到重要的数据必须多份存在.所以分布式就 ...

  4. 线上服务内存OOM问题定位三板斧

    相信大家都有感触,线上服务内存OOM的问题,是最难定位的问题,不过归根结底,最常见的原因: 本身资源不够 申请的太多 资源耗尽 58到家架构部,运维部,58速运技术部联合进行了一次线上服务内存OOM问 ...

  5. python三大神器之装饰器

    装饰器的形成过程 假如你要写一个计算函数执行时间的函数,代码如下: import time def func1(): print('in func1') def timer(func): def in ...

  6. R语言爬虫:CSS方法与XPath方法对比(表格介绍)

    css 选择器与 xpath 用法对比 目标 匹配节点 CSS 3 XPath 所有节点 ~ * //* 查找一级.二级.三级标题节点 <h1>,<h2>,<h3> ...

  7. HBase简介(梳理知识)

    一. 简介 hbase是bigtable的开源山寨版本.是建立的hdfs之上,提供高可靠性.高性能.列存储.可伸缩.实时读写的数据库系统.它介于nosql和RDBMS之间,仅能通过主键(row key ...

  8. 前端页面加载速度优化---Ngnix之GZIP压缩

    gzip on; #开启Gzip gzip_static on;#是否开启gzip静态资源 #nginx对于静态文件的处理模块,该模块可以读取预先压缩的gz文件,这样可以减少每次请求进行gzip压缩的 ...

  9. 北京Uber优步司机奖励政策(3月3日)

    滴快车单单2.5倍,注册地址:http://www.udache.com/ 如何注册Uber司机(全国版最新最详细注册流程)/月入2万/不用抢单:http://www.cnblogs.com/mfry ...

  10. 北京Uber优步司机奖励政策(12月2日)

    用户组:人民优步(适用于12月2日)奖励政策: 滴快车单单2.5倍,注册地址:http://www.udache.com/ 如何注册Uber司机(全国版最新最详细注册流程)/月入2万/不用抢单:htt ...