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SHU 2013 暑期集训（7-18）解题报告

Problem A (hdu 1086)

You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5559 Accepted Submission(s): 2660

Problem Description

Many geometry（几何）problems were designed in the ACM/ICPC. And now, I also prepare a geometry problem for this final exam. According to the experience of many ACMers, geometry problems are always much trouble, but this problem is very easy, after all we are now attending an exam, not a contest :)
Give you N (1<=N<=100) segments（线段）, please output the number of all intersections（交点）. You should count repeatedly if M (M>2) segments intersect at the same point.
Note:
You can assume that two segments would not intersect at more than one point.

Input

Input contains multiple test cases. Each test case contains a integer N (1=N<=100) in a line first, and then N lines follow. Each line describes one segment with four float values x1, y1, x2, y2 which are coordinates of the segment’s ending.
A test case starting with 0 terminates the input and this test case is not to be processed.

Output

For each case, print the number of intersections, and one line one case.

Sample Input

0.00 0.00 1.00 1.00

0.00 1.00 1.00 0.00

0.00 0.00 1.00 1.00

0.00 1.00 1.00 0.000

0.00 0.00 1.00 0.00 0

Sample Output

1 3

Author

lcy

题目大意：给你n个线段，让你求这些线段有多少个交点。

分析:线段相交的判断，使用快速排斥试验和跨立试验，快速排斥实验即计算两个线段为对角线的两个矩形是否有重合部分，跨立实验则是分别检测一个线段的两个点是不是在另一个线段的两侧（叉积符号相反）。

  1 //============================================================================
  2 // Name        : 基本函数模板.cpp
  3 // Author      :
  4 // Version     :
  5 // Copyright   : Your copyright notice
  6 // Description : Hello World in C++, Ansi-style
  7 //============================================================================
  8 #include <iostream>
  9 #include <stdio.h>
 10 #include <string.h>
 11 #include <algorithm>
 12 #include <queue>
 13 #include <map>
 14 #include <vector>
 15 #include <set>
 16 #include <string>
 17 #include <math.h>
 18 #define maxlen 110
 19 using namespace std;
 20 
 21 const double eps = 1e-8;
 22 const double PI = acos(-1.0);
 23 int sgn(double x)
 24 {
 25     if(fabs(x) < eps)return 0;
 26     if(x < 0)return -1;
 27     else return 1;
 28 }
 29 struct Point
 30 {
 31     double x,y;
 32     Point(){}
 33     Point(double _x,double _y)
 34     {
 35         x = _x;y = _y;
 36     }
 37     Point operator -(const Point &b)const
 38     {
 39         return Point(x - b.x,y - b.y);
 40     }
 41     //叉积
 42     double operator ^(const Point &b)const
 43     {
 44         return x*b.y - y*b.x;
 45     }
 46     //点积
 47     double operator *(const Point &b)const
 48     {
 49         return x*b.x + y*b.y;
 50     }
 51     //绕原点旋转角度B（弧度值），后x,y的变化
 52     void transXY(double B)
 53     {
 54         double tx = x,ty = y;
 55         x = tx*cos(B) - ty*sin(B);
 56         y = tx*sin(B) + ty*cos(B);
 57     }
 58 };
 59 struct Line
 60 {
 61     Point s,e;
 62     Line(){}
 63     Line(Point _s,Point _e)
 64     {
 65         s = _s;e = _e;
 66     }
 67     //两直线相交求交点
 68     //第一个值为0表示直线重合，为1表示平行，为0表示相交,为2是相交
 69     //只有第一个值为2时，交点才有意义
 70     pair<int,Point> operator &(const Line &b)const
 71     {
 72         Point res = s;
 73         if(sgn((s-e)^(b.s-b.e)) == 0)
 74         {
 75             if(sgn((s-b.e)^(b.s-b.e)) == 0)
 76                 return make_pair(0,res);//重合
 77             else return make_pair(1,res);//平行
 78         }
 79         double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
 80         res.x += (e.x-s.x)*t;
 81         res.y += (e.y-s.y)*t;
 82         return make_pair(2,res);
 83     }
 84 };
 85 //*两点间距离
 86 double dist(Point a,Point b)
 87 {
 88     return sqrt((a-b)*(a-b));
 89 }
 90 //*判断线段相交
 91 bool inter(Line l1,Line l2)
 92 {
 93     return
 94     max(l1.s.x,l1.e.x) >= min(l2.s.x,l2.e.x) &&
 95     max(l2.s.x,l2.e.x) >= min(l1.s.x,l1.e.x) &&
 96     max(l1.s.y,l1.e.y) >= min(l2.s.y,l2.e.y) &&
 97     max(l2.s.y,l2.e.y) >= min(l1.s.y,l1.e.y) &&
 98     sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e)) <= 0 &&
 99     sgn((l1.s-l2.e)^(l2.s-l2.e))*sgn((l1.e-l2.e)^(l2.s-l2.e)) <= 0;
100 }
101 //判断直线和线段相交
102 bool Seg_inter_line(Line l1,Line l2) //判断直线l1和线段l2是否相交
103 {
104     return sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e)) <= 0;
105 }
106 //点到直线距离
107 //返回为result,是点到直线最近的点
108 Point PointToLine(Point P,Line L)
109 {
110     Point result;
111     double t = ((P-L.s)*(L.e-L.s))/((L.e-L.s)*(L.e-L.s));
112     result.x = L.s.x + (L.e.x-L.s.x)*t;
113     result.y = L.s.y + (L.e.y-L.s.y)*t;
114     return result;
115 }
116 //点到线段的距离
117 //返回点到线段最近的点
118 Point NearestPointToLineSeg(Point P,Line L)
119 {
120     Point result;
121     double t = ((P-L.s)*(L.e-L.s))/((L.e-L.s)*(L.e-L.s));
122     if(t >= 0 && t <= 1)
123     {
124         result.x = L.s.x + (L.e.x - L.s.x)*t;
125         result.y = L.s.y + (L.e.y - L.s.y)*t;
126     }
127     else
128     {
129         if(dist(P,L.s) < dist(P,L.e))
130             result = L.s;
131         else result = L.e;
132     }
133     return result;
134 }
135 //计算多边形面积
136 //点的编号从0~n-1
137 double CalcArea(Point p[],int n)
138 {
139     double res = 0;
140     for(int i = 0;i < n;i++)
141         res += (p[i]^p[(i+1)%n])/2;
142     return fabs(res);
143 }
144 //*判断点在线段上
145 bool OnSeg(Point P,Line L)
146 {
147     return
148     sgn((L.s-P)^(L.e-P)) == 0 &&
149     sgn((P.x - L.s.x) * (P.x - L.e.x)) <= 0 &&
150     sgn((P.y - L.s.y) * (P.y - L.e.y)) <= 0;
151 }
152 int n;
153 Line a[maxlen];
154 int main()
155 {
156     while(scanf("%d",&n)!=EOF)
157     {
158         if(n==0)
159             break;
160         for(int i=0;i<n;++i)
161         {
162             scanf("%lf%lf%lf%lf",&a[i].s.x,&a[i].s.y,&a[i].e.x,&a[i].e.y);
163         }
164         int cnt=0;
165         for(int i=0;i<n;++i)
166         {
167             for(int j=i+1;j<n;++j)
168             {
169                 if(inter(a[i],a[j]))
170                     cnt++;
171             }
172         }
173         printf("%d
",cnt);
174     }
175     return 0;
176 }

View Code

Problem C(poj 1296)

Intersecting Lines

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8450		Accepted: 3840

Description

We all know that a pair of distinct points on a plane defines a line and that a pair of lines on a plane will intersect in one of three ways: 1) no intersection because they are parallel, 2) intersect in a line because they are on top of one another (i.e. they are the same line), 3) intersect in a point. In this problem you will use your algebraic knowledge to create a program that determines how and where two lines intersect.
Your program will repeatedly read in four points that define two lines in the x-y plane and determine how and where the lines intersect. All numbers required by this problem will be reasonable, say between -1000 and 1000.

Input

The first line contains an integer N between 1 and 10 describing how many pairs of lines are represented. The next N lines will each contain eight integers. These integers represent the coordinates of four points on the plane in the order x1y1x2y2x3y3x4y4. Thus each of these input lines represents two lines on the plane: the line through (x1,y1) and (x2,y2) and the line through (x3,y3) and (x4,y4). The point (x1,y1) is always distinct from (x2,y2). Likewise with (x3,y3) and (x4,y4).

Output

There should be N+2 lines of output. The first line of output should read INTERSECTING LINES OUTPUT. There will then be one line of output for each pair of planar lines represented by a line of input, describing how the lines intersect: none, line, or point. If the intersection is a point then your program should output the x and y coordinates of the point, correct to two decimal places. The final line of output should read "END OF OUTPUT".

Sample Input

5
0 0 4 4 0 4 4 0
5 0 7 6 1 0 2 3
5 0 7 6 3 -6 4 -3
2 0 2 27 1 5 18 5
0 3 4 0 1 2 2 5

Sample Output

INTERSECTING LINES OUTPUT
POINT 2.00 2.00
NONE
LINE
POINT 2.00 5.00
POINT 1.07 2.20
END OF OUTPUT

Source

Mid-Atlantic 1996

题目大意：判断两个线段的关系，（相交、平行、重合）如果相交，求出交点。

分析：模板题，判断平行很简单，就是向量平行，如果有有一个点在另一个线段上那么就是重合。

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 using namespace std;
 5 struct point
 6 {
 7     double  x,y;
 8 };
 9 point mid;
10 int Inter(point a,point b,point c,point d)//平行返回0 重合返回1 相交返回2
11 {
12     double A1=b.y-a.y,A2=d.y-c.y,B1=a.x-b.x,B2=c.x-d.x;
13     double C1=b.y*(b.x-a.x)-b.x*(b.y-a.y),C2=d.y*(d.x-c.x)-d.x*(d.y-c.y);
14     if(A1*B2==B1*A2)
15     {
16         if(A2*C1==A1*C2&&B1*C2==B2*C1)
17             return 1;
18         return 0;
19     }
20     mid.x=(B1*C2-B2*C1)/(B2*A1-B1*A2);
21     mid.y=(A1*C2-A2*C1)/(A2*B1-A1*B2);
22     return 2;
23 }
24 int main()
25 {
26     point a,b,c,d;
27     int n;
28     while(scanf("%d",&n)!=EOF)
29     {
30         puts("INTERSECTING LINES OUTPUT");
31         while(n--)
32         {
33             scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y,&d.x,&d.y);
34             int ans=Inter(a,b,c,d);
35             if(ans==0)
36                 printf("NONE
");
37             else if(ans==1)
38                 printf("LINE
");
39             else
40                 printf("POINT %.2f %.2f
",mid.x,mid.y);
41         }
42         puts("END OF OUTPUT");
43     }
44     return 0;
45 }

View Code

Problem D (poj 2653)

Pick-up sticks

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 7807		Accepted: 2881

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

Sample Output

Top sticks: 2, 4, 5.
Top sticks: 1, 2, 3.

Hint

Huge input,scanf is recommended.

Source

Waterloo local 2005.09.17

题目分析：给你n个线段，问你有多少根线段的顶上没有线段与其相交，输出这些线段的序号。

分析：还是线段相交问题，枚举两个线段，如果有相交，那么标记，最后输出没有被标记的线段即可。

  1 #include <iostream>
  2 #include <stdio.h>
  3 #include <string.h>
  4 #include <algorithm>
  5 #include <queue>
  6 #include <map>
  7 #include <vector>
  8 #include <set>
  9 #include <string>
 10 #include <math.h>
 11 using namespace std;
 12 
 13 const double eps = 1e-8;
 14 const double PI = acos(-1.0);
 15 int sgn(double  x)
 16 {
 17     if(fabs(x) < eps)return 0;
 18     if(x < 0)return -1;
 19     else return 1;
 20 }
 21 struct Point
 22 {
 23     double   x,y;
 24     Point(){}
 25     Point(double  _x,double   _y)
 26     {
 27         x = _x;y = _y;
 28     }
 29     Point operator -(const Point &b)const
 30     {
 31         return Point(x - b.x,y - b.y);
 32     }
 33     //叉积
 34     double operator ^(const Point &b)const
 35     {
 36         return x*b.y - y*b.x;
 37     }
 38     //点积
 39     double operator *(const Point &b)const
 40     {
 41         return x*b.x + y*b.y;
 42     }
 43     //绕原点旋转角度B（弧度值），后x,y的变化
 44     void transXY(double B)
 45     {
 46         double tx = x,ty = y;
 47         x = tx*cos(B) - ty*sin(B);
 48         y = tx*sin(B) + ty*cos(B);
 49     }
 50 };
 51 struct Line
 52 {
 53     Point s,e;
 54     Line(){}
 55     Line(Point _s,Point _e)
 56     {
 57         s = _s;e = _e;
 58     }
 59 };
 60 //*两点间距离
 61 double dist(Point a,Point b)
 62 {
 63     return sqrt((a-b)*(a-b));
 64 }
 65 //*判断线段相交
 66 bool inter(Line l1,Line l2)
 67 {
 68     return
 69     max(l1.s.x,l1.e.x) >= min(l2.s.x,l2.e.x) &&
 70     max(l2.s.x,l2.e.x) >= min(l1.s.x,l1.e.x) &&
 71     max(l1.s.y,l1.e.y) >= min(l2.s.y,l2.e.y) &&
 72     max(l2.s.y,l2.e.y) >= min(l1.s.y,l1.e.y) &&
 73     sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e)) <= 0 &&
 74     sgn((l1.s-l2.e)^(l2.s-l2.e))*sgn((l1.e-l2.e)^(l2.s-l2.e)) <= 0;
 75 }
 76 const int MAXN = 100010;
 77 Line l[MAXN];
 78 bool f[MAXN];
 79 int n;
 80 int main ()
 81 {
 82     while(scanf("%d",&n)!=EOF)
 83     {
 84         if(n==0)
 85             break;
 86         for(int i=0;i<n;++i)
 87         {
 88             scanf("%lf%lf%lf%lf",&l[i].s.x,&l[i].s.y,&l[i].e.x,&l[i].e.y);
 89         }
 90         memset(f,0,sizeof(f));
 91         for(int i=0;i<n;++i)
 92         {
 93             for(int j=i+1;j<n;++j)
 94             {
 95                 if(inter(l[i],l[j]))
 96                 {
 97                     f[i]=1;
 98                     break;
 99                 }
100             }
101         }
102         printf("Top sticks:");
103         for(int i=0;i<n;++i)
104         {
105             if(f[i]==0)
106             {
107                 if(i!=n-1)
108                 printf(" %d,",i+1);
109                 else
110                     printf(" %d.
",i+1);
111             }
112         }
113     }
114 
115 }

View Code

Problem G (poj 2954)

Triangle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4394		Accepted: 1942

Description

A lattice point is an ordered pair (x, y) where x and y are both integers. Given the coordinates of the vertices of a triangle (which happen to be lattice points), you are to count the number of lattice points which lie completely inside of the triangle (points on the edges or vertices of the triangle do not count).

Input

The input test file will contain multiple test cases. Each input test case consists of six integers x₁, y₁, x₂, y₂, x₃, and y₃, where (x₁, y₁), (x₂, y₂), and (x₃, y₃) are the coordinates of vertices of the triangle. All triangles in the input will be non-degenerate (will have positive area), and −15000 ≤ x₁, y₁, x₂, y₂, x₃, y₃ ≤ 15000. The end-of-file is marked by a test case with x₁ = y₁ = x₂ = y₂ = x₃ = y₃ = 0 and should not be processed.

Output

For each input case, the program should print the number of internal lattice points on a single line.

Sample Input

0 0 1 0 0 1
0 0 5 0 0 5
0 0 0 0 0 0

Sample Output

0
6

Source

Stanford Local 2004

题目大意：给你三角型的三个点的坐标（整数），问你在这个三角形内部有多少个整点（横纵坐标都为整数）.

分析：

线段上的格点数算法：包括端点的整点个数为gcd(abs(x1-x2),abs(y1-y2))+1个

Pick公式：整点多边形的面积=内部整点个数+边上的整点个数/2-1.

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <string.h>
 4 #include <algorithm>
 5 #include <queue>
 6 #include <map>
 7 #include <vector>
 8 #include <set>
 9 #include <string>
10 #include <math.h>
11 using namespace std;
12 
13 const double eps = 1e-8;
14 const double PI = acos(-1.0);
15 int sgn(double x)
16 {
17     if(fabs(x) < eps)return 0;
18     if(x < 0)return -1;
19     else return 1;
20 }
21 struct Point
22 {
23     int x,y;
24     Point(){}
25     Point(int  _x,int  _y)
26     {
27         x = _x;y = _y;
28     }
29     Point operator -(const Point &b)const
30     {
31         return Point(x - b.x,y - b.y);
32     }
33     //叉积
34     double operator ^(const Point &b)const
35     {
36         return x*b.y - y*b.x;
37     }
38     //点积
39     double operator *(const Point &b)const
40     {
41         return x*b.x + y*b.y;
42     }
43     //绕原点旋转角度B（弧度值），后x,y的变化
44     void transXY(double B)
45     {
46         double tx = x,ty = y;
47         x = tx*cos(B) - ty*sin(B);
48         y = tx*sin(B) + ty*cos(B);
49     }
50 };
51 int gcd(int a,int b)
52 {
53     if(b==0)
54         return a;
55     return gcd(b,a%b);
56 }
57 int myabs(int a)
58 {
59     return a<0?-a:a;
60 }
61 int main()
62 {
63     Point a,b,c;
64     while(scanf("%d%d%d%d%d%d",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y)!=EOF)
65     {
66         if(a.x==0&&a.y==0&&b.x==0&&b.y==0&&c.x==0&&c.y==0)
67             break;
68         int s=myabs((b-a)^(c-a));
69         int e=gcd(abs(a.x-b.x),abs(a.y-b.y))+gcd(abs(a.x-c.x),abs(a.y-c.y))+gcd(abs(c.x-b.x),abs(c.y-b.y));
70         int ans=(s-e+2)/2;
71         printf("%d
",ans);
72     }
73 }

View Code

Problem H (poj 2954)

Shape of HDU

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4276 Accepted Submission(s): 1899

Problem Description

话说上回讲到海东集团推选老总的事情，最终的结果是XHD以微弱优势当选，从此以后，“徐队”的称呼逐渐被“徐总”所取代，海东集团（HDU）也算是名副其实了。
创业是需要地盘的，HDU向钱江肉丝高新技术开发区申请一块用地，很快得到了批复，据说这是因为他们公司研发的“海东牌”老鼠药科技含量很高，预期将占全球一半以上的市场。政府划拨的这块用地是一个多边形，为了描述它，我们用逆时针方向的顶点序列来表示，我们很想了解这块地的基本情况，现在请你编程判断HDU的用地是凸多边形还是凹多边形呢？

Input

输入包含多组测试数据，每组数据占2行，首先一行是一个整数n，表示多边形顶点的个数，然后一行是2×n个整数，表示逆时针顺序的n个顶点的坐标（xi,yi），n为0的时候结束输入。

Output

对于每个测试实例，如果地块的形状为凸多边形，请输出“convex”,否则输出”concave”，每个实例的输出占一行。

Sample Input

0 0

1 0

1 1

0 1

Sample Output

convex 海东集团终于顺利成立了！后面的路，他们会顺顺利利吗？欲知后事如何，且听下回分解——

Author

lcy

Source

ACM程序设计_期末考试（时间已定！！）

Recommend

lcy

题目大意：问n多边形是否为凸多边形。

分析:凸多边形的判断就是将相邻的两个边的是否向一个方向旋转。

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <algorithm>
 5 #include <queue>
 6 #include <map>
 7 #include <vector>
 8 #include <set>
 9 #include <string>
10 #include <math.h>
11 #define maxlen 1010
12 using namespace std;
13 
14 const double eps = 1e-8;
15 const double PI = acos(-1.0);
16 int sgn(double x)
17 {
18     if(fabs(x) < eps)return 0;
19     if(x < 0)return -1;
20     else return 1;
21 }
22 struct Point
23 {
24     double x,y;
25     Point() {}
26     Point(double _x,double _y)
27     {
28         x = _x;
29         y = _y;
30     }
31     Point operator -(const Point &b)const
32     {
33         return Point(x - b.x,y - b.y);
34     }
35     //叉积
36     double operator ^(const Point &b)const
37     {
38         return x*b.y - y*b.x;
39     }
40     //点积
41     double operator *(const Point &b)const
42     {
43         return x*b.x + y*b.y;
44     }
45     //绕原点旋转角度B（弧度值），后x,y的变化
46     void transXY(double B)
47     {
48         double tx = x,ty = y;
49         x = tx*cos(B) - ty*sin(B);
50         y = tx*sin(B) + ty*cos(B);
51     }
52 };
53 bool isconvex(Point poly[],int n)
54 {
55     //bool s[3];
56     //memset(s,false,sizeof(s));
57     for(int i = 0; i < n; i++)
58     {
59         double temp=(poly[(i+1)%n].x-poly[i].x)*(poly[(i+2)%n].y-poly[i].y) - (poly[(i+2)%n].x-poly[i].x)*(poly[(i+1)%n].y-poly[i].y);
60         if(temp<0)
61         {
62             return false;
63         }
64     }
65     return true;
66 }
67 Point p[maxlen];
68 int n;
69 int main()
70 {
71     while(scanf("%d",&n)!=EOF)
72     {
73         if(n==0)
74             break;
75         for(int i=0; i<n; ++i)
76             scanf("%d%d",&p[i].x,&p[i].y);
77         if(isconvex(p,n))
78             printf("convex
");
79         else
80             printf("concave
");
81     }
82     return 0;
83 }

View Code

Problem H (hdu 1392)

Surround the Trees

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5995 Accepted Submission(s): 2256

Problem Description

There are a lot of trees in an area. A peasant wants to buy a rope to surround all these trees. So at first he must know the minimal required length of the rope. However, he does not know how to calculate it. Can you help him?
The diameter and length of the trees are omitted, which means a tree can be seen as a point. The thickness of the rope is also omitted which means a rope can be seen as a line.

There are no more than 100 trees.

Input

The input contains one or more data sets. At first line of each input data set is number of trees in this data set, it is followed by series of coordinates of the trees. Each coordinate is a positive integer pair, and each integer is less than 32767. Each pair is separated by blank.

Zero at line for number of trees terminates the input for your program.

Output

The minimal length of the rope. The precision should be 10^-2.

Sample Input

9 12 7 24 9 30 5 41 9 80 7 50 87 22 9 45 1 50 7 0

Sample Output

243.06

Source

Asia 1997, Shanghai (Mainland China)

Recommend

Ignatius.L

题目大意：有n个柱子，你要将他们都围起来，问你需要绳子的最小长度。

分析：凸包问题，Graham算法。主要过程就是先选取最左下角的点作为p0，然后按极角逆时针排序编号（p1-pn-1），现将p0,p1入栈，每次判断下一个点的时候，将其与栈顶的两个点做叉积，看是否符号相反，若相反则将栈顶元素退出，知道相同为止。重复这个操作就可以将外围的点存入在栈中。

这题要求求的是最短，即求栈中相邻点距离之和。

注意：如果栈中只有两个点，那么答案就是两者的距离。

  1 #include <iostream>
  2 #include <stdio.h>
  3 #include <string.h>
  4 #include <algorithm>
  5 #include <math.h>
  6 #include <iomanip>
  7 using namespace std;
  8 
  9 const double eps = 1e-8;
 10 const double PI = acos(-1.0);
 11 int sgn(int x)
 12 {
 13     if(fabs(double (x)) < eps)return 0;
 14     if(x < 0)return -1;
 15     else return 1;
 16 }
 17 struct Point
 18 {
 19     int   x,y;
 20     Point() {}
 21     Point(int _x,int  _y)
 22     {
 23         x = _x;
 24         y = _y;
 25     }
 26     Point operator -(const Point &b)const
 27     {
 28         return Point(x-b.x,y-b.y);
 29     }
 30     double operator ^(const Point &b)const
 31     {
 32         return x*b.y - y*b.x;
 33     }
 34     double operator *(const Point &b)const
 35     {
 36         return x*b.x + y*b.y;
 37     }
 38 };
 39 //*两点间距离
 40 double dist(Point a,Point b)
 41 {
 42     return sqrt((a-b)*(a-b));
 43 }
 44 
 45 const int MAXN = 1010;
 46 Point list[MAXN];
 47 int Stack[MAXN],top;
 48 //相对于list[0]的极角排序
 49 bool cmp(Point p1,Point p2)
 50 {
 51     double tmp = (p1-list[0])^(p2-list[0]);
 52     if(sgn(tmp) > 0)return true;
 53     else if(sgn(tmp) == 0 && sgn(dist(p1,list[0]) - dist(p2,list[0])) <= 0)
 54         return true;
 55     else return false;
 56 }
 57 void Graham(int n)
 58 {
 59     Point p0;
 60     int k = 0;
 61     p0 = list[0];
 62     //找最下边的一个点
 63     for(int i = 1; i < n; i++)
 64     {
 65         if( (p0.y > list[i].y) || (p0.y == list[i].y && p0.x > list[i].x) )
 66         {
 67             p0 = list[i];
 68             k = i;
 69         }
 70     }
 71     swap(list[k],list[0]);
 72     sort(list+1,list+n,cmp);
 73     if(n == 1)
 74     {
 75         top = 1;
 76         Stack[0] = 0;
 77         return;
 78     }
 79     if(n == 2)
 80     {
 81         top = 2;
 82         Stack[0] = 0;
 83         Stack[1] = 1;
 84         return ;
 85     }
 86     Stack[0] = 0;
 87     Stack[1] = 1;
 88     top = 2;
 89     for(int i = 2; i < n; i++)
 90     {
 91         while(top > 1 && sgn((list[Stack[top-1]]-list[Stack[top-2]])^(list[i]-list[Stack[top-2]])) <= 0)
 92             top--;
 93         Stack[top++] = i;
 94     }
 95 }
 96 int n;
 97 int main()
 98 {
 99     while(scanf("%d",&n)!=EOF)
100     {
101         if(n==0)
102             break;
103         for(int i=0; i<n; ++i)
104             scanf("%d%d",&list[i].x,&list[i].y);
105         Graham(n);
106         double ans=0.0;
107         if (top==2)
108             ans+=dist(list[0],list[1]);
109         else
110         {
111             for(int i=0; i<top; ++i)
112             {
113                 ans+=dist(list[Stack[i]],list[Stack[(i+1)%top]]);
114             }
115         }
116         printf("%.2f
",ans);
117     }
118     return 0;
119 }

View Code

感谢kuangbin学长的模板，还是很好用的，不会的等填坑。╮(╯▽╰)╭

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原文地址：https://www.cnblogs.com/shuzy/p/3199666.html