经典算法详解（10）图中有多少个三角形

题目：请说出下面图形中包含多少个三角形？请用一个程序完成计算。

C++版本

#include<iostream>

using namespace std;

const char NO_POINT = '0';

//任意的一条线
const char *map[] = { "ad","ab","db","ae","aj","ah","ej","eh","jh","af","ak","ai","fk","fi","ki","ag","ac","gc",
"de","df","dg","ef","eg","fg","bj","bk","bg","jk","jg","kg","bh","bi","bc","hi","hc","ic" };
//共线的点
const char *line[] = { "adb","aejh","afki","agc","defg","bjkg","bhic" };

//点是否在线上
int contain( const char *str, char a) {
    int i = 0;
    while (str[i] != '') {        //注意字符使用单引号，字符串是双引号
        if (str[i] == a)
            return 1;
        i++;
    }
    return 0;
}

//三个点是否在一条线上函数
int isInALine(const char *str[], char a, char b, char c) {
    int i ;
    for (i = 0; i < 7; i++) {
        if (contain(str[i], a) && contain(str[i], b) && contain(str[i], c)) {
            return 1;
        }
    }
    return 0;
}

//两条线的交点函数
char getCrossPoint(const char *str1, const char *str2) {
    if (*str1 == *str2)
        return *str1;
    if (*str1 == *(str2 + 1))
        return *str1;
    if (*(str1 + 1) == *str2)
        return *(str1 + 1);
    if (*(str1 + 1) == *(str2 + 1))
        return *(str1 + 1);
    return NO_POINT;
}

//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。
int isTriangle(const char *str1, const char *str2, const char *str3) {
    char Point1, Point2, Point3;
    Point1 = getCrossPoint(str1, str2);
    if (Point1 == NO_POINT)
        return 0;
    Point2 = getCrossPoint(str1, str3);
    if (Point2 == NO_POINT)
        return 0;
    Point3 = getCrossPoint(str2, str3);
    if (Point3 == NO_POINT)
        return 0;
    if (isInALine(line, Point1, Point2, Point3))
        return 0;
    return 1;
}

int getTriangelCount( const char *str[]) {
    int i, j, k,count=0;
    for (i = 0; i < 36; i++) {
        for (j = i+1; j < 36; j++) {
            for (k = j+1; k < 36; k++) {
                if (isTriangle(str[i], str[j], str[k]))
                    count++;
            }
        }
    }
    return count;
}

int main(int argc, char *argv[]) {
    cout << getTriangelCount(map);
    getchar();
    return 0;
}

 解题思路：

（1）给每个交点做标记，如下：

（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：

（3）故需要有如下一些子函数：求两条线的交点，三个点是否共线等。