python/TangDou/LuoGuBook/ZhiMuSanJiaoXing.cpp

#include <bits/stdc++.h>
const int N = 110;
char a[N][N];
/*
4
abccddadca

2
aaa
*/
using namespace std;

struct Node {
    int x, y;
    char c;
};
vector<Node> q;
/*
把半径当做 1 ，建立坐标系，然后枚举就行了，需要注意的是，在判断三条边的长度是否相等时用 double 牵扯到精度问题，
解决方法是:比较长度的平方，长度的平方肯定是整数，还有就是层层之间纵坐标相差是sqrt(3) 的倍数，
为了不牵扯到小数，把纵坐标的 sqrt(3) 当做1 ，求边长时乘三即可。
*/
int pf(int x) {
    return x * x;
}

bool check(Node a, Node b, Node c) {
    cout << "a.x=" << a.x << ",a.y=" << a.y << endl;
    cout << "b.x=" << b.x << ",b.y=" << a.y << endl;
    cout << "c.x=" << c.x << ",c.y=" << a.y << endl;

    // a与b之间距离 不等于 a与c之间距离
    if (pf(a.x - b.x) + pf(a.y - b.y) * 3 != pf(a.x - c.x) + pf(a.y - c.y) * 3) return false;
    // a与b之间距离 不等于 b与c之间距离
    if (pf(a.x - b.x) + pf(a.y - b.y) * 3 != pf(b.x - c.x) + pf(b.y - c.y) * 3) return false;

    return true;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("ZhiMuSanJiaoXing.in", "r", stdin);
#endif
    int n;
    cin >> n;
    string s;
    cin >> s;
    // 先把字符串存入char[][]
    int idx = 0;
    for (int i = 0; i < n; i++)      // n行
        for (int j = 0; j <= i; j++) // i列
            a[i][j] = s[idx++];

    // 把所有坐标记录下来
    for (int i = 0; i < n; i++) {
        for (int j = 0; j <= i; j++) {
            q.push_back({n - 1 - i, j, a[i][j]});
            // cout << "x=" << n - 1 - i << ",y=" << j << " " << a[i][j] << endl;
        }
    }

    vector<char> res;
    // 在q数组中，选择任意三个，计算两两间距离是不是相等
    for (int i = 0; i < q.size(); i++)
        for (int j = 0; j < q.size(); j++)
            for (int k = 0; k < q.size(); k++) {
                if (i == j || i == k || j == k) continue;

                if (q[i].c != q[j].c || q[i].c != q[k].c || q[j].c != q[k].c) continue;
                if (check(q[i], q[j], q[k]))
                    res.push_back(q[i].c);
            }

    if (res.size() == 0)
        cout << "No Solution" << endl;
    else {
        sort(res.begin(), res.end());
        for (char x : res) cout << x << endl;
    }
    return 0;
}