POJ_2653_Pick-up sticks_判斷線段相交
阿新 • • 發佈：2018-05-10
esp 一個 沒有 for each truct fine soft ini jpg
The picture to the right below illustrates the first case from input.
POJ_2653_Pick-up sticks_判斷線段相交
Description
Stan has n sticks of various length. He throws them one at a time on the floor in a random way. After finishing throwing, Stan tries to find the top sticks, that is these sticks such that there is no stick on top of them. Stan has noticed that the last thrown stick is always on top but he wants to know all the sticks that are on top. Stan sticks are very, very thin such that their thickness can be neglected.Input
Input consists of a number of cases. The data for each case start with 1 <= n <= 100000, the number of sticks for this case. The following n lines contain four numbers each, these numbers are the planar coordinates of the endpoints of one stick. The sticks are listed in the order in which Stan has thrown them. You may assume that there are no more than 1000 top sticks. The input is ended by the case with n=0. This case should not be processed.Output
For each input case, print one line of output listing the top sticks in the format given in the sample. The top sticks should be listed in order in which they were thrown.The picture to the right below illustrates the first case from input.
Sample Input
5 1 1 4 2 2 3 3 1 1 -2.0 8 4 1 4 8 2 3 3 6 -2.0 3 0 0 1 1 1 0 2 1 2 0 3 1 0
Sample Output
Top sticks: 2, 4, 5.
Top sticks: 1, 2, 3.
斯坦有各種長度的n條。他在地板上隨意地扔了一個。在完成投擲後，斯坦試圖找到最上面的棍子，那就是這些棍子，這樣就沒有棍子在上面了。
斯坦註意到，最後一根投擲棒總是在上面，但他想知道上面所有的棍子。斯坦棒非常非常薄，以至於它們的厚度可以被忽略。
暴力可過的一道題。直接枚舉所有的所有的線段判斷能不能被後面的覆蓋即可。
然後判斷線段相交用四次叉積判斷即可。
代碼：
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <math.h>
using namespace std;
typedef double f2;
#define N 100050
#define eps 1e-6
bool vis[N];
int ans[N],n;
struct Point {
f2 x,y;
Point() {}
Point(f2 x_,f2 y_) :
x(x_),y(y_) {}
Point operator + (const Point &p) const {return Point(x+p.x,y+p.y);}
Point operator - (const Point &p) const {return Point(x-p.x,y-p.y);}
Point operator * (f2 rate) const {return Point(x*rate,y*rate);}
};
f2 dot(const Point &p1,const Point &p2) {return p1.x*p2.x+p1.y*p2.y;}
f2 cross(const Point &p1,const Point &p2) {return p1.x*p2.y-p1.y*p2.x;}
f2 FABS(f2 x) {return x>0?x:-x;}
struct Line {
Point p,v;
Line() {}
Line(const Point &p_,const Point &v_) :
p(p_),v(v_) {}
};
Line a[N];
f2 turn(const Point &p1,const Point &p2,const Point &p3) {
return cross(p3-p1,p2-p1);
}
bool judge(const Line &l1,const Line &l2) {
if(turn(l1.p,l1.v,l2.p)*turn(l1.p,l1.v,l2.v)>0) return 0;
if(turn(l2.p,l2.v,l1.p)*turn(l2.p,l2.v,l1.v)>0) return 0;
return 1;
}
void init() {
memset(vis,0,sizeof(vis)); ans[0]=0;
}
void solve() {
init();
int i,j;
f2 x,y,z,w;
int fir=0;
for(i=1;i<=n;i++) {
scanf("%lf%lf%lf%lf",&a[i].p.x,&a[i].p.y,&a[i].v.x,&a[i].v.y);
}
printf("Top sticks:");
for(i=1;i<=n;i++) {
int flg=0;
for(j=i+1;j<=n;j++) {
if(judge(a[i],a[j])) {
flg=1; break;
}
}
if(!flg) {
if(!fir) {
fir=1;
}else printf(",");
printf(" %d",i);
}
}
puts(".");
}
int main() {
while(scanf("%d",&n)&&n) {
solve();
}
}
POJ_2653_Pick-up sticks_判斷線段相交