POJ 2036 I Conduit!

 

I Conduit!

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 1313		Accepted: 245

Description

Irv Kenneth Diggit works for a company that excavates trenches, digs holes and generally tears up people's yards. Irv's job is to make sure that no underground pipe or cable is underneath where excavation is planned. He has several different maps, one for each utility company, showing where their conduits lie, and he needs to draw one large, consolidated map combining them all. One approach would be to simply draw each of the smaller maps one at a time onto the large map. However, this often wastes time, not to mention ink for the pen-plotter in the office, since in many cases portions of the conduits overlap with each other (albeit at different depths underground). What Irv wants is a way to determine the minimum number of line segments to draw given all the line segments from the separate maps.

Input

Input will consist of multiple input sets. Each set will start with a single line containing a positive integer n indicating the total number of line segments from all the smaller maps. Each of the next n lines will contain a description of one segment in the format

x1 y1 x2 y2

where (x1,y1) are the coordinates of one endpoint and (x2,y2) are the coordinates of the other. Coordinate values are floating point values in the range 0...1000 speci.ed to at most two decimal places. The maximum number of line segments will be 10000 and all segments will have non-zero length. Following the last input set there will be a line containing a 0 indicating end of input; it should not be processed.

Output

For each input set, output on a single line the minimum number of line segments that need to be drawn on the larger, consolidated map.

Sample Input

3
1.0 10.0 3.0 14.0
0.0 0.0 20.0 20.0
10.0 28.0 2.0 12.0
2
0.0 0.0 1.0 1.0
1.0 1.0 2.15 2.15
2
0.0 0.0 1.0 1.0
1.0 1.0 2.15 2.16
0

Sample Output

2
1
2

Source

East Central North America 2004

 

 

/* 呵呵，這題進前20了 13 6262545 bobten2008 732K 141MS C++ 2246B 2009-12-20 14:52:15 很噁心的計算幾何問題,核心思想是利用排序將seg按照斜率進行排序,容易錯誤的地方有四點 1)精度問題,需要加eps修正. 2)輸入數據中的每個seg的兩個點的順序不是按照從左到右或者從下到上,因此需要調整seg的 兩個端點比如對於線短1 2 3 2,也有可能以3 2 1 2的形式出現,我在這裏統一調成從下到上, 從左到右的順序,這裏如果不調整可能會影響後續的工作. 3)排序算法的比較函數,比較複雜,需要考慮豎線(即沒有斜率的情況),詳見正文裏的註釋 4)最後順序掃描排序後的seg時,需要調整當前seg的x2或者y2值,比如對於 [1 2 10 2], [2 2 5 2] [6 2 12 2]這組數據,排序完也是這個順序,那麼第二個seg被包含在第一個 seg裏,如果不修正第二個seg的y2座標,那麼在處理第三個seg時,第三個seg肯定不被包含進來,那麼 最後答案是2,其實這組數據的答案應該是1.所以在處理完第二個seg後需要將其y2修正爲10. 具體怎麼修正呢,很簡單,只要取y2爲當前seg的y2和上一個seg的y2的最大值即可 算法的時間複雜度是O(nlgn) */ #include <iostream> #include <algorithm> #include <cmath> #define EPS 1e-8 #define MAX_N 10005 using namespace std; struct seg { double x1, y1, x2, y2; bool spe; //spe = true表示當前seg與y軸平行,否則不平行,與y軸平行時是沒有斜率的 double a, b; }segs[MAX_N]; int n; //比較函數,核心思想是把沒有斜率(與y軸平行)的放在有斜率的後面,把斜率小的放在斜率大的前面 //對於具有相同斜率的seg,把與y軸交點低的放在前面 //對於在一條線上的seg,把x座標(當與y軸平行時比較y座標)小的放在前面 bool compare(const seg &seg1, const seg &seg2) { //兩個seg都沒有斜率 if(seg1.spe && seg2.spe) return fabs(seg1.x1 - seg2.x1) < EPS ? seg1.y1 < seg2.y1 + EPS : seg1.x1 < seg2.x2 + EPS; else if(seg1.spe && !seg2.spe) return false; //一個有斜率一個沒有斜率,把有斜率的放在前面 else if(!seg1.spe && seg2.spe) return true; else //兩個seg都有斜率 { //斜率相等時需要比較與y軸的交點,當焦點也相等時需要比較左端點x值的大小 if(fabs(seg1.a - seg2.a) < EPS) return fabs(seg1.b - seg2.b) < EPS ? seg1.x1 < seg2.x1 + EPS : seg1.b < seg2.b + EPS; else return seg1.a < seg2.a + EPS; } } void swap(double &d1, double &d2) { double temp = d1; d1 = d2; d2 = temp; } //判斷兩個seg是否相交, type用來區分這個兩個seg與y軸的平行情況,type = 0時表示與y軸平行 bool cross(const seg& seg1, const seg &seg2, bool type) { if(!type) return seg2.y1 <= seg1.y2; else return seg2.x1 <= seg1.x2; } int main() { int i; while(scanf("%d", &n) && n != 0) { for(i = 0; i < n; i++) { scanf("%lf%lf%lf%lf", &segs[i].x1, &segs[i].y1, &segs[i].x2, &segs[i].y2); if(segs[i].x1 > segs[i].x2 + EPS) { swap(segs[i].x1, segs[i].x2); swap(segs[i].y1, segs[i].y2);} else if(fabs(segs[i].x1 - segs[i].x2) < EPS && segs[i].y1 > segs[i].y2 + EPS) { swap(segs[i].x1, segs[i].x2); swap(segs[i].y1, segs[i].y2); } segs[i].spe = false; if(fabs(segs[i].x1 - segs[i].x2) < EPS) segs[i].spe = true; else { segs[i].a = (segs[i].y2 - segs[i].y1) / (segs[i].x2 - segs[i].x1); segs[i].b = segs[i].y1 - segs[i].x1 * segs[i].a; } } sort(segs, segs + n, compare); int countv = 1; for(i = 1; i < n; i++) { //當前seg與y軸平行其與上一個seg相交 if(segs[i].spe && segs[i - 1].spe && fabs(segs[i].x1 - segs[i - 1].x1) < EPS && cross(segs[i - 1], segs[i], false)) { //調整當前seg的y2 if(segs[i].y2 < segs[i - 1].y2 + EPS) segs[i].y2 = segs[i - 1].y2; continue; } //當前seg不與y軸平行且與上一個seg相交 if(!segs[i].spe && !segs[i - 1].spe && fabs(segs[i].a - segs[i - 1].a) < EPS && fabs(segs[i].b - segs[i - 1].b) < EPS && cross(segs[i - 1], segs[i], true)) { //調整當前seg的x2 if(segs[i].x2 < segs[i - 1].x2 + EPS) segs[i].x2 = segs[i - 1].x2; continue; } //找到一個新的不可與上一個seg合併的seg則需要爲countv的值增加1 countv++; } printf("%d/n", countv); } return 0; }