1 #include<iostream>
2
3 using namespace std;
4
5 const char NO_POINT = '0';
6
7 //任意的一条线
8 const char *map[] = { "ad","ab","db","ae","aj","ah","ej","eh","jh","af","ak","ai","fk","fi","ki","ag","ac","gc",
9 "de","df","dg","ef","eg","fg","bj","bk","bg","jk","jg","kg","bh","bi","bc","hi","hc","ic" };
10 //共线的点
11 const char *line[] = { "adb","aejh","afki","agc","defg","bjkg","bhic" };
12
13 //点是否在线上
14 int contain( const char *str, char a) {
15 int i = 0;
16 while (str[i] != '\0') { //注意字符使用单引号,字符串是双引号
17 if (str[i] == a)
18 return 1;
19 i++;
20 }
21 return 0;
22 }
23
24 //三个点是否在一条线上函数
25 int isInALine(const char *str[], char a, char b, char c) {
26 int i ;
27 for (i = 0; i < 7; i++) {
28 if (contain(str[i], a) && contain(str[i], b) && contain(str[i], c)) {
29 return 1;
30 }
31 }
32 return 0;
33 }
34
35 //两条线的交点函数
36 char getCrossPoint(const char *str1, const char *str2) {
37 if (*str1 == *str2)
38 return *str1;
39 if (*str1 == *(str2 + 1))
40 return *str1;
41 if (*(str1 + 1) == *str2)
42 return *(str1 + 1);
43 if (*(str1 + 1) == *(str2 + 1))
44 return *(str1 + 1);
45 return NO_POINT;
46 }
47
48 //三条线两两必须有交点,并且三条线不能共线才能构成三角形。
49 int isTriangle(const char *str1, const char *str2, const char *str3) {
50 char Point1, Point2, Point3;
51 Point1 = getCrossPoint(str1, str2);
52 if (Point1 == NO_POINT)
53 return 0;
54 Point2 = getCrossPoint(str1, str3);
55 if (Point2 == NO_POINT)
56 return 0;
57 Point3 = getCrossPoint(str2, str3);
58 if (Point3 == NO_POINT)
59 return 0;
60 if (isInALine(line, Point1, Point2, Point3))
61 return 0;
62 return 1;
63 }
64
65 int getTriangelCount( const char *str[]) {
66 int i, j, k,count=0;
67 for (i = 0; i < 36; i++) {
68 for (j = i+1; j < 36; j++) {
69 for (k = j+1; k < 36; k++) {
70 if (isTriangle(str[i], str[j], str[k]))
71 count++;
72 }
73 }
74 }
75 return count;
76 }
77
78 int main(int argc, char *argv[]) {
79 cout << getTriangelCount(map);
80 getchar();
81 return 0;
82 }
(2)总共有36条线段,如果三条线段两两之间存在交点,但一条线上(已经包含了三条线交于同一点),则可以构成三角形。如下图所示,最左边的构成三角形,右边两个不构成三角形:
![]()