POJ 计算几何专项训练（1） 【2318】&【2398】&【3304】&【2653】&【1556】&【1066】
POJ 2318 TOYS
题意是在一个大矩形里有n条分割线把矩形分割成n+1部分、再给出一些玩具的坐标、要求统计每个部分内有多少个玩具、
具体做法就是二分求解出当前玩具右方的第一条线、这可以用叉积判断、
Code:
var
res:array [0..5002] of longint;
p:array [0..5002] of record x1,x2,y1,y2:double;end;
xx,yy,ldx,ldy,rux,ruy:double;
n,m,i,l,r,mid:longint;
function crossp(x1,y1,x2,y2:double):double;
begin
crossp:=x1*y2-x2*y1;
end;
function ok(now:longint):boolean;
begin
if crossp(p[now].x1-xx,p[now].y1-yy,p[now].x2-xx,p[now].y2-yy)>0 then exit(true) else exit(false);
end;
begin
read(n);
while n<>0 do
begin
readln(m,ldx,ruy,rux,ldy);
for i:=1 to n do
begin
readln(p[i].x2,p[i].x1);
p[i].y1:=ldy;
p[i].y2:=ruy;
end;
p[0].x1:=ldx;p[0].x2:=ldx;
p[0].y1:=ldy;p[0].y2:=rux;
p[n+1].x1:=rux;p[n+1].x2:=rux;
p[n+1].y1:=ldy;p[n+1].y2:=ruy;
for i:=1 to m do
begin
readln(xx,yy);
l:=1;r:=n+1;
while l<r-1 do
begin
mid:=(l+r) div 2;
if ok(mid) then r:=mid else l:=mid;
end;
while (l>1) and ok(l-1) do dec(l);
while not ok(l) do inc(l);
inc(res[l-1]);
end;
for i:=0 to n do
writeln(i,': ',res[i]);
writeln;
fillchar(res,sizeof(res),0);
read(n);
end;
end.
POJ 2398 TOY Storage
本题是上题的姊妹题，区别在于本题要求统计的是包含X个玩具的部分有多少、
P.S 上题的X坐标是有序给出的，而本题要先排序、
Code:
var
res,app:array [0..1002] of longint;
p:array [0..1002] of record x1,x2,y1,y2:double;end;
xx,yy,ldx,ldy,rux,ruy:double;
n,m,i,l,r,mid:longint;
function crossp(x1,y1,x2,y2:double):double;
begin
crossp:=x1*y2-x2*y1;
end;
function ok(now:longint):boolean;
begin
if crossp(p[now].x1-xx,p[now].y1-yy,p[now].x2-xx,p[now].y2-yy)>0 then exit(true) else exit(false);
end;
procedure sort(l,r:longint);
var
i,j:longint;
cx:double;
begin
i:=l;j:=r;cx:=p[random(r-l+1)+l].x1;
repeat
while p[i].x1<cx do inc(i);
while cx<p[j].x1 do dec(j);
if i<=j then
begin
p[1002]:=p[i];p[i]:=p[j];p[j]:=p[1002];
inc(i);dec(j);
end;
until i>j;
if i<r then sort(i,r);
if j>l then sort(l,j);
end;
begin
read(n);
while n<>0 do
begin
readln(m,ldx,ruy,rux,ldy);
for i:=1 to n do
begin
readln(p[i].x2,p[i].x1);
p[i].y1:=ldy;
p[i].y2:=ruy;
end;
sort(1,n);
p[n+1].x1:=rux;p[n+1].x2:=rux;
p[n+1].y1:=ldy;p[n+1].y2:=ruy;
for i:=1 to m do
begin
readln(xx,yy);
l:=1;r:=n+1;
while l<r-1 do
begin
mid:=(l+r) div 2;
if ok(mid) then r:=mid else l:=mid;
end;
while (l>1) and ok(l-1) do dec(l);
while not ok(l) do inc(l);
inc(res[l-1]);
end;
for i:=0 to n do
inc(app[res[i]]);
writeln('Box');
for i:=1 to n do
if app[i]<>0 then
writeln(i,': ',app[i]);
fillchar(app,sizeof(app),0);
fillchar(res,sizeof(res),0);
read(n);
end;
end.
POJ 3304 Segments
本题要求判断是否存在一条直线通过给出的所有直线、
不难想到如果该直线不过任何一条直线的端点，一定可以让它旋转一定角度与某一个端点相碰；
而如果该直线只与某一个相碰，必然可以旋转另一端点与另一个端点相碰、
于是，我们N^2枚举出直线，然后判断是否与每一条线段相交即可、
Code:
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#define EPS 1e-8
#define sqr(x) ((x)*(x))
using namespace std;
double x[501],y[501];
int n,vv;
double dis(double aa,double bb,double cc,double dd){
return sqrt((aa-bb)*(aa-bb)+(cc-dd)*(cc-dd));
}
double crossp(double aa,double bb,double cc,double dd){
return aa*dd-bb*cc;
}
int judge(double x1,double y1,double x2,double y2){
if (dis(x1,x2,y1,y2)<EPS) return 0;
for (int j=1;j<=n*2;j++)
if (j&1)
if (crossp(x[j]-x1,y[j]-y1,x2-x1,y2-y1)*crossp(x2-x1,y2-y1,x[j+1]-x1,y[j+1]-y1)<0) return 0;
return 1;
}
int main(){
scanf("%d",&vv);
while (vv--){
scanf("%d",&n);
for (int i=1;i<=n*2;i++)
scanf("%lf%lf",&x[i],&y[i]);
if (n==1){
printf("Yes!\n");
continue;
}
int succ=0;
for (int i=1;i<=n*2 && !succ;i++)
for (int j=i+1;j<=n*2 && !succ;j++)
if (judge(x[i],y[i],x[j],y[j])) succ++;
if (succ) printf("Yes!\n"); else printf("No!\n");
}
}
POJ 1269 Pick-up Sticks
本题来自于一个经典的……游戏？
题意主要就是每次放一条线段上去、如果后放的与先放的有交点就覆盖掉先放的、
于是我们可以维护一个随便什么链表、然后维护当前在顶上的线段编号、因为题目保证TOP线段不超过1000条、所以是可以无压力通过的、、
Code:
#include <cstring>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#define EPS 1e-9
#define sqr(x) ((x)*(x))
#define INF 1e9
#define pb push_back
using namespace std;
struct segment{
double sx,sy,ex,ey;
}ss[100010];
int n,top[1010],tt;
double crossp(double x1,double y1,double x2,double y2){
return x1*y2-x2*y1;
}
int cross(segment a,segment b){
if (max(a.sx,a.ex)<min(b.sx,b.ex) || max(a.sy,a.ey)<min(b.sy,b.ey) \
|| min(a.sx,a.ex)>max(b.sx,b.ex) || min(a.sy,a.ey)>max(b.sy,b.ey)) return 0;
if (crossp(a.sx-b.sx,a.sy-b.sy,b.ex-b.sx,b.ey-b.sy)*crossp(b.ex-b.sx,b.ey-b.sy,a.ex-b.sx,a.ey-b.sy)>0 && \
crossp(b.sx-a.sx,b.sy-a.sy,a.ex-a.sx,a.ey-a.sy)*crossp(a.ex-a.sx,a.ey-a.sy,b.ex-a.sx,b.ey-a.sy)>0) return 1;
else return 0;
}
int main(){
scanf("%d",&n);
while (n){
for (int i=1;i<=n;i++)
scanf("%lf%lf%lf%lf",&ss[i].sx,&ss[i].sy,&ss[i].ex,&ss[i].ey);
top[tt=1]=1;
for (int i=2;i<=n;i++){
for (int j=1;j<=tt;j++)
if (cross(ss[top[j]],ss[i])) top[j]=-1;
int cur=0;
for (int j=1;j<=tt;j++)
if (top[j]>0) top[++cur]=top[j];
tt=cur;
top[++tt]=i;
}
printf("Top sticks:");
for (int i=1;i<tt;i++)
printf(" %d,",top[i]);
printf(" %d.\n",top[tt]);
scanf("%d",&n);
}
return 0;
}
POJ 1556 The doors
一看图就知道是什么题了、
把所有出现的点全部建成一张图，然后求解最短路就可以了。
Code:
#include <cstring>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#define EPS 1e-9
#define sqr(x) ((x)*(x))
#define INF 1e9
#define pb push_back
using namespace std;
struct segment{
double sx,sy,ex,ey;
}ss[100010];
int n,top[1010],tt;
double crossp(double x1,double y1,double x2,double y2){
return x1*y2-x2*y1;
}
int cross(segment a,segment b){
if (max(a.sx,a.ex)<min(b.sx,b.ex) || max(a.sy,a.ey)<min(b.sy,b.ey) \
|| min(a.sx,a.ex)>max(b.sx,b.ex) || min(a.sy,a.ey)>max(b.sy,b.ey)) return 0;
if (crossp(a.sx-b.sx,a.sy-b.sy,b.ex-b.sx,b.ey-b.sy)*crossp(b.ex-b.sx,b.ey-b.sy,a.ex-b.sx,a.ey-b.sy)>0 && \
crossp(b.sx-a.sx,b.sy-a.sy,a.ex-a.sx,a.ey-a.sy)*crossp(a.ex-a.sx,a.ey-a.sy,b.ex-a.sx,b.ey-a.sy)>0) return 1;
else return 0;
}
int main(){
scanf("%d",&n);
while (n){
for (int i=1;i<=n;i++)
scanf("%lf%lf%lf%lf",&ss[i].sx,&ss[i].sy,&ss[i].ex,&ss[i].ey);
top[tt=1]=1;
for (int i=2;i<=n;i++){
for (int j=1;j<=tt;j++)
if (cross(ss[top[j]],ss[i])) top[j]=-1;
int cur=0;
for (int j=1;j<=tt;j++)
if (top[j]>0) top[++cur]=top[j];
tt=cur;
top[++tt]=i;
}
printf("Top sticks:");
for (int i=1;i<tt;i++)
printf(" %d,",top[i]);
printf(" %d.\n",top[tt]);
scanf("%d",&n);
}
return 0;
}
POJ 1066 Treasure Hunt
本题大概是说、在（0，0）（100，100）的矩形内有若干线段把矩形分割成若干区域，然后给出一个宝藏的所在地（X,Y）求一条从边界进入到达该点的路径，使得该路径通过的线段最少、
本题与3304有一定相似之处，不难发现只要枚举所有线段的端点到（X,Y）组成的线段与其它线段的交点数就可以了、
Code:
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
using namespace std;
struct segment{
double beginx,beginy,endx,endy;
}ss[1010];
int n;
double nx,ny;
segment temp;
double crossp(double x1,double y1,double x2,double y2){
return x1*y2-x2*y1;
}
int cross(segment a,segment b){
if (min(a.beginx,a.endx)>max(b.beginx,b.endx) ||\
min(b.beginx,b.endx)>max(a.beginx,a.endx) ||\
min(a.beginy,a.endy)>max(b.beginy,b.endy) ||\
min(b.beginy,b.endy)>max(a.beginy,a.endy)) return 0;
if (crossp(a.beginx-b.beginx,a.beginy-b.beginy,b.endx-b.beginx,b.endy-b.beginy)\
*crossp(b.endx-b.beginx,b.endy-b.beginy,a.endx-b.beginx,a.endy-b.beginy)>0 &&\
crossp(b.beginx-a.beginx,b.beginy-a.beginy,a.endx-a.beginx,a.endy-a.beginy)\
*crossp(a.endx-a.beginx,a.endy-a.beginy,b.endx-a.beginx,b.endy-a.beginy)>0) return 1;
return 0;
}
int main(){
scanf("%d",&n);
for (int i=1;i<=n;i++)
scanf("%lf%lf%lf%lf",&ss[i].beginx,&ss[i].beginy,&ss[i].endx,&ss[i].endy);
scanf("%lf%lf",&nx,&ny);
temp.endx=nx;temp.endy=ny;
int ans=99999999;
for (int i=1;i<=n;i++){
int cnt=0;
temp.beginx=ss[i].beginx;
temp.beginy=ss[i].beginy;
for (int j=1;j<=n;j++)
if (cross(ss[j],temp)) cnt++;
if (cnt<ans) ans=cnt;
cnt=0;
temp.beginx=ss[i].endx;
temp.beginy=ss[i].endy;
for (int j=1;j<=n;j++)
if (cross(ss[j],temp)) cnt++;
if (cnt<ans) ans=cnt;
}
if (ans==99999999) ans=0;
printf("Number of doors = %d\n",ans+1);
return 0;
}
