Ultimate Weapon
| Time Limit: 2000MS | Memory Limit: 131072K | |
| Total Submissions: 2430 | Accepted: 1173 |
Description
In year 2008 of the Cosmic Calendar, the Aliens send a huge armada towards the Earth seeking after conquest. The humans now depend on their ultimate weapon to retain their last hope of survival. The weapon, while capable of creating a continuous, closed and convex lethal region in the space and annihilating everything enclosed within, unfortunately exhausts upon each launch a tremendous amount of energy which is proportional to the surface area of the lethal region.
Given the positions of all battleships in the Aliens' armada, your task is to calculate the minimum amount of energy required to destroy the armada with a single launch of the ultimate weapon. You need to report the surface area of the lethal region only.
Input
The first line contains one number N -- the number of battleships.(1 ≤ N ≤ 500)
Following N lines each contains three integers presenting the position of one battleship.
Output
The minimal area rounded to three decimal places.
Sample Input
4
0 0 0
4 0 0
2 3 0
1 1 2
Sample Output
19.137
Hint
There are no four coplaner battleships.
Source
POJ Founder Monthly Contest – 2008.03.16, Jiang Liyang
三维凸包裸题，求这些点组成的凸包的表面积，数据最大为500，用增量算法，且由于只求表面积，可以不用合并同一平面的三角形。
1 #include<iostream>
2 #include<algorithm>
3 #include<cmath>
4 #include<cstdio>
5 using namespace std;
6 const int N = 510;
7 const double eps = 1e-8;
8 typedef struct point3 {
9 double x, y, z;
10 point3() { }
11 point3(double a, double b, double c) :x(a), y(b), z(c) { }
12 point3 operator -(const point3 &b)const { //返回减去后的新点
13 return point3(x - b.x, y - b.y, z - b.z);
14 }
15 point3 operator +(const point3 &b)const { //返回加上后的新点
16 return point3(x + b.x, y + b.y, z + b.z);
17 }
18 //数乘计算
19 point3 operator *(const double &k)const { //返回相乘后的新点
20 return point3(x * k, y * k, z*k);
21 }
22 point3 operator /(const double &k)const { //返回相除后的新点
23 return point3(x / k, y / k, z / k);
24 }
25 double operator *(const point3 &b)const { //点乘
26 return (x*b.x + y*b.y + z*b.z);
27 }
28 point3 operator ^(const point3 &p) const { //叉积
29 return point3(y*p.z - p.y*z, z*p.x - x*p.z, x*p.y - y*p.x);
30 }
31 double vlen()const { //向量的模
32 return sqrt(x*x + y*y + z*z);
33 }
34 }point3;
35 struct fac {
36 int a, b, c;//凸包一个面上的三个点的编号
37 bool ok; //该面是否是最终凸包中的面
38 };
39 struct T3dhull {
40 int n; //初始点数
41 point3 ply[N]; //初始点
42 int trianglecnt; //凸包上三角形数
43 fac tri[N]; //凸包三角形可证明被创建的面不会超过6N
44 int vis[N][N]; //点i到点j是属于哪个面
45 double dist(point3 a) { return sqrt(a.x*a.x + a.y*a.y + a.z*a.z); } //两点长度
46 double area(point3 a, point3 b, point3 c){return dist((b - a) ^ (c - a));} //三角形面积*2
47
48 //返回四面体有向体积*6
49 //在储存面时，保证面的法线方向朝向凸包外部，如果在某一平面和点p所组成的四面体的有向体积为正，则p点在凸包外部，并且此点可以被p点看见。
50 double volume(point3 a, point3 b, point3 c, point3 d){
51 return ((b - a) ^ (c - a))* (d - a);
52 }
53 double ptoplane(point3 &p, fac &f){ //点到平面距离，体积法
54 point3 m = ply[f.b] - ply[f.a], n = ply[f.c] - ply[f.a], t = p - ply[f.a];
55 return (m^n) * t;
56 }
57 void deal(int p, int a, int b) {
58 int f = vis[a][b];
59 fac add;
60 if (tri[f].ok)
61 {
62 if ((ptoplane(ply[p], tri[f])) > eps)
63 dfs(p, f);
64 else
65 {
66 add.a = b, add.b = a, add.c = p, add.ok = 1;
67 vis[p][b] = vis[a][p] = vis[b][a] = trianglecnt;
68 tri[trianglecnt++] = add;
69 }
70 }
71 }
72 void dfs(int p, int cnt) {//维护凸包，如果点p在凸包外侧则更新凸包
73 tri[cnt].ok = 0;
74 deal(p, tri[cnt].b, tri[cnt].a);
75 deal(p, tri[cnt].c, tri[cnt].b);
76 deal(p, tri[cnt].a, tri[cnt].c);
77 }
78 bool same(int s, int e) {
79 point3 a = ply[tri[s].a], b = ply[tri[s].b], c = ply[tri[s].c];
80 return fabs(volume(a, b, c, ply[tri[e].a])) < eps
81 && fabs(volume(a, b, c, ply[tri[e].b])) < eps
82 && fabs(volume(a, b, c, ply[tri[e].c])) < eps;
83 }
84 void construct()//构造凸包
85 {
86 int i, j;
87 trianglecnt = 0;
88 if (n<4) return;
89 bool tmp = true;
90 for (i = 1; i < n; i++) //前两点不共点
91 {
92 if ((dist(ply[0] - ply[i])) > eps)
93 {
94 swap(ply[1], ply[i]);
95 tmp = false;
96 break;
97 }
98 }
99 if (tmp)return;
100 tmp = true;
101 for (i = 2; i < n; i++) //前三点不共线
102 {
103 if ((dist((ply[0] - ply[1]) ^ (ply[1] - ply[i]))) > eps)
104 {
105 swap(ply[2], ply[i]);
106 tmp = false;
107 break;
108 }
109 }
110 if (tmp) return;
111 tmp = true;
112 for (i = 3; i < n; i++) //前四点不共面
113 {
114 if (fabs(((ply[0] - ply[1]) ^ (ply[1] - ply[2]))* (ply[0] - ply[i]))>eps)
115 {
116 swap(ply[3], ply[i]);
117 tmp = false;
118 break;
119 }
120 }
121 if (tmp)return;
122 fac add;
123 for (i = 0; i < 4; i++) //构建初始四面体
124 {
125 add.a = (i + 1) % 4, add.b = (i + 2) % 4, add.c = (i + 3) % 4, add.ok = 1;
126 if ((ptoplane(ply[i], add))>0)
127 swap(add.b, add.c);
128 vis[add.a][add.b] = vis[add.b][add.c] = vis[add.c][add.a] = trianglecnt;
129 tri[trianglecnt++] = add;
130 }
131 for (i = 4; i < n; i++) //构建更新凸包
132 {
133 for (j = 0; j < trianglecnt; j++)
134 {
135 if (tri[j].ok && (ptoplane(ply[i], tri[j])) > eps)
136 {
137 dfs(i, j); break;
138 }
139 }
140 }
141 int cnt = trianglecnt;
142 trianglecnt = 0;
143 for (i = 0; i < cnt; i++)
144 {
145 if (tri[i].ok)
146 tri[trianglecnt++] = tri[i];
147 }
148 }
149 double area() //表面积
150 {
151 double ret = 0;
152 for (int i = 0; i < trianglecnt; i++)
153 ret += area(ply[tri[i].a], ply[tri[i].b], ply[tri[i].c]);
154 return ret / 2.0;
155 }
156 double volume()
157 {
158 point3 p(0, 0, 0);
159 double ret = 0;
160 for (int i = 0; i < trianglecnt; i++)
161 ret += volume(p, ply[tri[i].a], ply[tri[i].b], ply[tri[i].c]);
162 return fabs(ret / 6);
163 }
164 }hull;
165
166 int main() {
167 while (~scanf("%d", &hull.n)) {
168 int i;
169 for (i = 0; i < hull.n; i++)
170 scanf("%lf %lf %lf", &hull.ply[i].x, &hull.ply[i].y, &hull.ply[i].z);
171 hull.construct();
172 printf("%.3lf
", hull.area());
173 }
174 return 0;
175 }