Hotter Colder
时间： 1ms 内存：64M
描述：
The children's game Hotter Colder is played as follows. Player A leaves the room while player B hides an object somewhere within. Player A re-enters at position (0, 0) and then visits various other locations about the room. When player A visits a new position, player B announces ``Hotter" if this position is closer to the object than the previous position, ``Colder" if it is farther, and ``Same" if it is the same distance.
输入：
Input consists of up to 50 lines, each containing an (x, y)-coordinate pair followed by ``Hotter'', ``Colder'', or ``Same''. Each pair represents a position within the room, which may be assumed to be a square with opposite corners at (0,0) and (10,10).
输出：
For each line of input, print a line giving the total area of the region in which the object may have been placed, to two decimal places. If there is no such region, output ``0.00''.
示例输入：
10.0 10.0 Colder
10.0 0.0 Hotter
0.0 0.0 Colder
10.0 10.0 Hotter
示例输出：
50.00
37.50
12.50
0.00
提示：
参考答案（内存最优[2160]）：
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <queue>
using namespace std;
#define INF 10000000
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
int dcmp(double x){
if(fabs(x)<eps)return 0;
return x>0?1:-1;
}
struct Point{
double x,y;
Point(double _x=0,double _y=0){
x=_x;y=_y;
}
};
Point operator + (const Point &a,const Point &b){
return Point(a.x+b.x,a.y+b.y);
}
Point operator - (const Point &a,const Point &b){
return Point(a.x-b.x,a.y-b.y);
}
Point operator * (const Point &a,const double &p){
return Point(a.x*p,a.y*p);
}
Point operator / (const Point &a,const double &p){
return Point(a.x/p,a.y/p);
}
bool operator < (const Point &a,const Point &b){
return a.x<b.x||(a.x==b.x&&a.y<b.y);
}
bool operator == (const Point &a,const Point &b){
return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
}
double Dot(Point a,Point b){
return a.x*b.x+a.y*b.y;
}
double Length(Point a){
return sqrt(Dot(a,a));
}
double Angle(Point a,Point b){
return acos(Dot(a,b)/Length(a)/Length(b));
}
double angle(Point a){
return atan2(a.y,a.x);
}
double Cross(Point a,Point b){
return a.x*b.y-a.y*b.x;
}
Point vecunit(Point a){
return a/Length(a);
}
Point Normal(Point a){
return Point(-a.y,a.x)/Length(a);
}
Point Rotate(Point a,double rad){
return Point(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));
}
double Area2(Point a,Point b,Point c){
return Length(Cross(b-a,c-a));
}
struct Line{
Point p,v;
double ang;
Line(){};
Line(Point p,Point v):p(p),v(v){
ang=atan2(v.y,v.x);
}
bool operator < (const Line &L) const {
return ang<L.ang;
}
};
bool OnLeft(const Line &L,const Point &p){
return dcmp(Cross(L.v,p-L.p))>=0;
}
Point GetLineIntersection(Point p,Point v,Point q,Point w){
Point u=p-q;
double t=Cross(w,u)/Cross(v,w);
return p+v*t;
}
Point GetLineIntersection(Line a,Line b){
return GetLineIntersection(a.p,a.v,b.p,b.v);
}
vector<Point> HPI(vector<Line> L){
int n=L.size();
sort(L.begin(),L.end());//将所有半平面按照极角排序。
int first,last;
vector<Point> p(n);
vector<Line> q(n);
vector<Point> ans;
q[first=last=0]=L[0];
for(int i=1;i<n;i++){
while(first<last&&!OnLeft(L[i],p[last-1]))last--;//删除顶部的半平面
while(first<last&&!OnLeft(L[i],p[first]))first++;//删除底部的半平面
q[++last]=L[i];//将当前的半平面假如双端队列顶部。
if(fabs(Cross(q[last].v,q[last-1].v))<eps){//对于极角相同的，选择性保留一个。
last--;
if(OnLeft(q[last],L[i].p))q[last]=L[i];
}
if(first<last)p[last-1]=GetLineIntersection(q[last-1],q[last]);//计算队列顶部半平面交点。
}
while(first<last&&!OnLeft(q[first],p[last-1]))last--;//删除队列顶部的无用半平面。
if(last-first<=1)return ans;//半平面退化
p[last]=GetLineIntersection(q[last],q[first]);//计算队列顶部与首部的交点。
for(int i=first;i<=last;i++)ans.push_back(p[i]);//将队列中的点复制。
return ans;
}
double PolyArea(vector<Point> p){
int n=p.size();
double ans=0;
for(int i=1;i<n-1;i++)
ans+=Cross(p[i]-p[0],p[i+1]-p[0]);
return fabs(ans)/2;
}
Point pp[200];
int main()
{
Point a,b;
vector<Line> L;
Line s;
a=Point(0,0);b=Point(10,0);s=Line(a,b-a);L.push_back(s);
a=Point(10,0);b=Point(10,10);s=Line(a,b-a);L.push_back(s);
a=Point(10,10);b=Point(0,10);s=Line(a,b-a);L.push_back(s);
a=Point(0,10);b=Point(0,0);s=Line(a,b-a);L.push_back(s);
Point pre,cur;
char str[100];
int flag=0;
while(~scanf("%lf%lf%s",&cur.x,&cur.y,str)){
if(flag)puts("0.00");
else{
if(str[0]=='S'){
flag=1;
puts("0.00");
}
else{
a=(pre+cur)/2;
b=Normal(cur-pre);
if(str[0]=='C')L.push_back(Line(a,b));
if(str[0]=='H')L.push_back(Line(a,Point(-b.x,-b.y)));
vector<Point> ans=HPI(L);
printf("%.2f\n",PolyArea(ans));
pre=cur;
}
}
}
return 0;
}
参考答案（时间最优[0]）：
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <queue>
using namespace std;
#define INF 10000000
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
int dcmp(double x){
if(fabs(x)<eps)return 0;
return x>0?1:-1;
}
struct Point{
double x,y;
Point(double _x=0,double _y=0){
x=_x;y=_y;
}
};
Point operator + (const Point &a,const Point &b){
return Point(a.x+b.x,a.y+b.y);
}
Point operator - (const Point &a,const Point &b){
return Point(a.x-b.x,a.y-b.y);
}
Point operator * (const Point &a,const double &p){
return Point(a.x*p,a.y*p);
}
Point operator / (const Point &a,const double &p){
return Point(a.x/p,a.y/p);
}
bool operator < (const Point &a,const Point &b){
return a.x<b.x||(a.x==b.x&&a.y<b.y);
}
bool operator == (const Point &a,const Point &b){
return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
}
double Dot(Point a,Point b){
return a.x*b.x+a.y*b.y;
}
double Length(Point a){
return sqrt(Dot(a,a));
}
double Angle(Point a,Point b){
return acos(Dot(a,b)/Length(a)/Length(b));
}
double angle(Point a){
return atan2(a.y,a.x);
}
double Cross(Point a,Point b){
return a.x*b.y-a.y*b.x;
}
Point vecunit(Point a){
return a/Length(a);
}
Point Normal(Point a){
return Point(-a.y,a.x)/Length(a);
}
Point Rotate(Point a,double rad){
return Point(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));
}
double Area2(Point a,Point b,Point c){
return Length(Cross(b-a,c-a));
}
struct Line{
Point p,v;
double ang;
Line(){};
Line(Point p,Point v):p(p),v(v){
ang=atan2(v.y,v.x);
}
bool operator < (const Line &L) const {
return ang<L.ang;
}
};
bool OnLeft(const Line &L,const Point &p){
return dcmp(Cross(L.v,p-L.p))>=0;
}
Point GetLineIntersection(Point p,Point v,Point q,Point w){
Point u=p-q;
double t=Cross(w,u)/Cross(v,w);
return p+v*t;
}
Point GetLineIntersection(Line a,Line b){
return GetLineIntersection(a.p,a.v,b.p,b.v);
}
vector<Point> HPI(vector<Line> L){
int n=L.size();
sort(L.begin(),L.end());//将所有半平面按照极角排序。
int first,last;
vector<Point> p(n);
vector<Line> q(n);
vector<Point> ans;
q[first=last=0]=L[0];
for(int i=1;i<n;i++){
while(first<last&&!OnLeft(L[i],p[last-1]))last--;//删除顶部的半平面
while(first<last&&!OnLeft(L[i],p[first]))first++;//删除底部的半平面
q[++last]=L[i];//将当前的半平面假如双端队列顶部。
if(fabs(Cross(q[last].v,q[last-1].v))<eps){//对于极角相同的，选择性保留一个。
last--;
if(OnLeft(q[last],L[i].p))q[last]=L[i];
}
if(first<last)p[last-1]=GetLineIntersection(q[last-1],q[last]);//计算队列顶部半平面交点。
}
while(first<last&&!OnLeft(q[first],p[last-1]))last--;//删除队列顶部的无用半平面。
if(last-first<=1)return ans;//半平面退化
p[last]=GetLineIntersection(q[last],q[first]);//计算队列顶部与首部的交点。
for(int i=first;i<=last;i++)ans.push_back(p[i]);//将队列中的点复制。
return ans;
}
double PolyArea(vector<Point> p){
int n=p.size();
double ans=0;
for(int i=1;i<n-1;i++)
ans+=Cross(p[i]-p[0],p[i+1]-p[0]);
return fabs(ans)/2;
}
Point pp[200];
int main()
{
Point a,b;
vector<Line> L;
Line s;
a=Point(0,0);b=Point(10,0);s=Line(a,b-a);L.push_back(s);
a=Point(10,0);b=Point(10,10);s=Line(a,b-a);L.push_back(s);
a=Point(10,10);b=Point(0,10);s=Line(a,b-a);L.push_back(s);
a=Point(0,10);b=Point(0,0);s=Line(a,b-a);L.push_back(s);
Point pre,cur;
char str[100];
int flag=0;
while(~scanf("%lf%lf%s",&cur.x,&cur.y,str)){
if(flag)puts("0.00");
else{
if(str[0]=='S'){
flag=1;
puts("0.00");
}
else{
a=(pre+cur)/2;
b=Normal(cur-pre);
if(str[0]=='C')L.push_back(Line(a,b));
if(str[0]=='H')L.push_back(Line(a,Point(-b.x,-b.y)));
vector<Point> ans=HPI(L);
printf("%.2f\n",PolyArea(ans));
pre=cur;
}
}
}
return 0;
}
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