这道题由于做了前面的四分树的题，思路不难想，就是需要注意的地方，两组数据之间要输出空行，最后后面不能输出，然后就是图形和序列的对应关系有两个特殊的，如果是一般图形，就按照刘汝佳说的对应，如果是全黑，那么对应于一个0，图形全黑，输出一个0，如果图形全黑，那么输入的也就只有一个0，图形全白，那么输入序列的时候输入是空的，只输入一个-1就结束了，如果需要我们输出序列我们也是输出空，一个换行都不要输出，这个是看题才知道了，刘汝佳就没有写

#include <bits/stdc++.h>#define fi first
#define se second
#define pb push_back
#define mk make_pair
#define sz(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()using namespace std;typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pa;char s[70][70];
vi v;
int dx[4] = {0, 0, 1, 1};
int dy[4] = {0, 1, 0, 1};void dfs(int r, int c, int len, string ss, int x) {ss += x + '0';int ok1 = 0, ok2 = 0;for (int i = r; i < r + len; i++) {for (int j = c; j < c + len; j++) {if (s[i][j] == '0') ok1 = 1;if (s[i][j] == '1') ok2 = 1;}}if (ok1 && ok2) {for (int i = 1; i <= 4; i++) {dfs(r + dx[i - 1] * len / 2, c + dy[i - 1] * len / 2, len / 2, ss, i);}} else if (!ok1 && ok2) {ss.erase(ss.begin());reverse(all(ss));int num = 0;for (int i = 0; i < sz(ss); i++) num = num * 5 + ss[i] - '0';v.pb(num);}
}void solve(int r, int c, int len, vector<char> vec) {if (sz(vec)) {if (vec[0] == '0') vec.erase(vec.begin());char ch = vec[0];vec.erase(vec.begin());solve(r + dx[ch - '1'] * len / 2, c + dy[ch - '1'] * len / 2, len / 2, vec);} else {for (int i = r; i < r + len; i++) {for (int j = c; j < c + len; j++) {s[i][j] = '*';}}}
}int main() {int n;int cnt = 1;while (cin >> n && n) {if (cnt != 1) cout << endl;cout << "Image " << cnt++ << endl;if (n > 0) {int ok1 = 0, ok2 = 0;for (int i = 0; i < n; i++) {cin >> s[i];for (int j = 0; j < n; j++) {if (s[i][j] == '0') ok1 = 1;if (s[i][j] == '1') ok2 = 1;}}if (ok1 && !ok2) cout << "Total number of black nodes = 0" << endl;if (!ok1 && ok2) cout << 0 << endl << "Total number of black nodes = 1" << endl;if (ok1 && ok2) {dfs(0, 0, n, "", 0);sort(all(v));int i = 0;while (i < sz(v)) {for (int j = 0; j < 12 && i < sz(v); i++, j++) {cout << v[i];if (j == 11 && i < sz(v) - 1 || i == sz(v) - 1) cout << endl;else cout << " ";}}cout << "Total number of black nodes = " << sz(v) << endl;}} else {int x;while (cin >> x && x != -1) {v.pb(x);}if (!sz(v)) {for (int i = 0; i < -n; i++) {for (int j = 0; j < -n; j++) {cout << '.' << (j < -n - 1 ? "" : "\n");}}} else if (sz(v) == 1 && v[0] == 0) {for (int i = 0; i < -n; i++) {for (int j = 0; j < -n; j++) {cout << '*' << (j < -n - 1 ? "" : "\n");}}} else {for (int i = 0; i < -n; i++) {for (int j = 0; j < -n; j++) {s[i][j] = '.';}}for (int i = 0; i < sz(v); i++) {vector<char> vec;vec.pb('0');while (v[i]) {vec.pb(v[i] % 5 + '0');v[i] /= 5;}solve(0, 0, -n, vec);}for (int i = 0; i < -n; i++) {for (int j = 0; j < -n; j++) {cout << s[i][j] << (j < -n - 1 ? "" : "\n");}}}}v.clear();}return 0;
}