2009年12月26日星期六.pku2165
计算几何
算法：将窗口投影到yz平面和xz平面，分别计算
上图是投影到yz平面的情况，可以算出斜率的交区间，如果不存在则无解，然后随便从中选出一个斜率。
然后算出和第一个窗口的Y交点
上图是投影到xz平面的情况，需要两两算出直线在x轴上的投影区间，然后随便选一个点作为出始点ansx。
然后由这个点开始，算出斜率的交区间，进而求出和第一条窗的x交点
有了这两个点，就可以利用比例，求出和其他所有窗的交点了
最后，很恶心的是精度，需要特别注意判断无解的情况，一定要用
int dcmp ( double x) { return (x > eps) - (x < -eps);}
我在精度上卡了好几次，最后double eps = 1e-10; 时过的
1 /*
2 * SOUR:pku2165
3 * ALGO:computational geometry
4 * DATE: 2009年 12月 26日 星期六 21:31:06 CST
5 * COMM:4
6 * */
7 #include<iostream>
8 #include<cstdio>
9 #include<cstdlib>
10 #include<cstring>
11 #include<algorithm>
12 #include<cassert>
13 #include<cmath>
14 using namespace std;
15 typedef long long LL;
16 const int maxint = 0x7fffffff;
17 const long long max64 = 0x7fffffffffffffffll;
18 const int N = 128;
19 int n;
20 double h,ansx,offset;
21 struct W {
22 double x1,y1;
23 double x2,y2,z;
24 }w[N];
25
26 double eps = 1e-10;
27 int dcmp(double x) { return (x > eps) - (x < -eps);}
28
29 const double inf = 1e30;
30
31 void ckmax(double &a,double b) { if(dcmp(a - b) < 0) a = b; }
32 void ckmin(double &a,double b) { if(dcmp(a - b) > 0) a = b; }
33
34 bool CalcY()
35 {
36 int i;
37 double up = inf,down = -inf;
38 for(i = 0;i < n;i++) {
39 ckmax(down,w[i].y1 / w[i].z);
40 ckmin(up,w[i].y2 / w[i].z);
41 }
42 if(dcmp(down - up) > 0) {
43 return false;
44 }
45 double tmp = (up + down)/ 2;
46 h = tmp * w[0].z;
47 //printf("CalcY succeded h = %.6f\n",h);
48 return true;
49 }
50
51 struct P
52 {
53 double x1,x2;
54 }interval[N*N];
55 int top;
56
57 double dist(double a,double b) { return sqrt(a * a + b * b); }
58
59 bool CalcX()
60 {
61 int i ,j;
62 double z,x,len,L;
63 top = 0;
64 for(i = 0;i < n;i++) {
65 for(j = i + 1;j < n;j++) {
66 z = w[j].z - w[i].z;
67 x = w[j].x2 - w[i].x1;
68 len = dist(z,x);
69 L = w[j].z * len / (w[j].z - w[i].z);
70 x = x/len * L;
71 interval[top].x1 = w[j].x2 - x;
72
73 z = w[j].z - w[i].z;
74 x = w[j].x1 - w[i].x2;
75 len = dist(z,x);
76 L = w[j].z * len / (w[j].z - w[i].z);
77 x = x/len * L;
78 interval[top].x2 = w[j].x1 - x;
79
80 top ++;
81 }
82 }
83 double left = -inf,right = inf;
84 for(i = 0;i < top;i++) {
85 //printf("[%d] x1= %.6f,x2 = %.6f\n",i,interval[i].x1,interval[i].x2);
86 ckmax(left,interval[i].x1);
87 ckmin(right,interval[i].x2);
88 }
89 //printf("left = %.6f,right = %.6f\n",left,right);
90 if(dcmp(left - right) > 0)
91 return false;
92 //if(left > right)
93 //return false;
94
95 ansx = (left + right) /2;
96 left = -inf,right = inf;
97 for(i = 0;i < n;i++) {
98 ckmax(left,(w[i].x1 - ansx) / w[i].z);
99 ckmin(right,(w[i].x2 - ansx) / w[i].z);
100 }
101 //assert(left < right);
102 double mid = (left + right)/2;
103 offset = mid * w[0].z;
104 return true;
105 }
106
107
108 int main()
109 {
110 int i,j,k;
111 scanf("%d",&n);
112 for(i = 0;i < n;i++) {
113 scanf("%lf%lf",&w[i].x1,&w[i].y1);
114 scanf("%lf%lf%lf",&w[i].x2,&w[i].y2,&w[i].z);
115 }
116 if(!CalcY()) {
117 printf("UNSOLVABLE\n");
118 return 0;
119 }
120 if(!CalcX()) {
121 printf("UNSOLVABLE\n");
122 return 0;
123 }
124
125 printf("SOLUTION\n");
126 printf("%.6f\n",ansx);
127 for(i = 0;i < n;i++) {
128 double x = w[i].z * offset / w[0].z + ansx;
129 double y = w[i].z * h / w[0].z;
130 double z = w[i].z;
131 printf("%.6f %.6f %.6f\n",x,y,z);
132 }
133 return 0;
134 }
135
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