图的哈希:
图中的每两个点两两计算欧几里得距离,再加和。则这个图不管怎么旋转,镜像,都会映射到同一个hash值。
用pair数组pair<int,int> q[]
去存下dfs得到的一个图的所有点
#include<bits/stdc++.h>
#define x first
#define y second
using namespace std;
const int N = 110;
const double eps = 1e-6;
char g[N][N];
typedef pair<int, int> PII;
PII q[N * N];
int n, m;
int top;
double get_dist(PII a, PII b) {
int dx = a.x - b.x;
int dy = a.y - b.y;
return sqrt(dx * dx + dy * dy);
}
double get_hash() {
double sum = 0;
for (int i = 0; i < top; i++) {
for (int j = i + 1; j < top; j++) {
sum += get_dist(q[i], q[j]);
}
}
return sum;
}
char get_id(double key) {
static double hash[N];
static int id = 0;
for (int i = 0; i < id; i++) {
if (fabs(key - hash[i]) < eps)
return i + 'a';
}
hash[id++] = key;
return id - 1 + 'a';
}
void dfs(int a, int b) {
q[top++] = {a, b};
g[a][b] = '0';
for (int i = a - 1; i <= a + 1; i++) {
for (int j = b - 1; j <= b + 1; j++) {
if (i == a && j == b)continue;
if (i >= 0 && i < n && j >= 0 && j < m && g[i][j] == '1')
dfs(i, j);
}
}
}
int main() {
cin >> m >> n;
for (int i = 0; i < n; i++)cin >> g[i];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (g[i][j] == '1') {
top = 0;
dfs(i, j);
char c = get_id(get_hash());
for (int i = 0; i < top; i++) {
int a = q[i].x, b = q[i].y;
g[a][b] = c;
}
}
}
}
for (int i = 0; i < n; i++) {
cout << g[i] << endl;
}
return 0;
}
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