Exercise 6-8 Spatial Structure UVa806

1. Title Description: Click to open the link

2. Problem-solving ideas: The subject to examine the four-minute tree traversal. The subject is known as a four-point tree for the number of black nodes; the number of known black nodes in turn constructs a four-minute tree. The first case is equivalent to coding, we use the first order traversal to solve. The Encode function is used to encode the entire four-minute tree, and if it is found that both 0 and 1 appear internally, it is divided into four parts, recursively encoded, and conversely, the color and path of the node are counted. This encodes the path into a decimal number for easy conversion. Similarly, if a black node is known, the path of each point is decoded first, and then the origin of the location is located along its path from the source of the coordinates, which is then marked as ' * '.

Note: To see the output format, for the first case, full 12 to wrap.

3. Code:

#define _crt_secure_no_warnings #include <iostream> #include <algorithm> #include <string> #include <sstream> #include <set> #include <vector> #include <stack> #include <map> #include < queue> #include <deque> #include <cstdlib> #include <cstdio> #include <cstring> #include < cmath> #include <ctime> #include <functional>using namespace Std;const int N = + 5;char G[n][n];vector <int>code;int n;struct node{int color;int dir;} node[n*n];void encode (int r, int c, int w, int dir, int &id)//represent: Start coordinate r, starting point coordinate C, Width w, encoded direction, node number {int OK = 1;char ch = g[r ][c];for (int i = R; I < r + W; i++) for (int j = C; J < C + W; + j) if (g[i][j]! = ch) {ok = 0; break;}  if (OK)//found a node {node[id].color = (ch = = ' 0 '? 0:1); int x = dir, k = 1;stack<int>s;while (x>0) {s.push (x%); x /= 10;} while (!s.empty ())//use a stack to flip 5 binary number to convert to 10 binary {node[id].dir + = S.top () *k; S.pop (); k *= 5;} id++;} else//recursive encoding {encode (R, C, W/2, dir * + 1, id); EncoDe (R, C + W/2, W/2, dir * + 2, id); Encode (R + W/2, C, W/2, dir * + 3, id); Encode (R + W/2, C + W/2, W/2 , dir * + 4, id);}} void Draw (int r, int c, int w) {for (int i = R; I < r + W; i++.) for (int j = C; J < C + W; b + +) g[i][j] = ' * ';} void Decode (vector<int> s) {int len = s.size (); for (int i = 0; i < len; i++) {int x = s[i], k = 1;node[i].color = 1 while (x>0)//decode the path, low is the lower path {node[i].dir + = (x% 5) *k;x/= 5, K *= 10;}}  for (int i = 0; i < len; i++) {int dir = Node[i].dir;int r = 0, c = 0, W = n;while (dir>0)//find the corresponding area along the path {int x = dir% 10;if (x = = 1) w/= 2;if (x = = 2) {c + = W/2; w/= 2;} if (x = = 3) {r + = W/2; w/= 2;} if (x = = 4) {r + = W/2; c + = W/2; w/= 2;} Dir/= 10;} Draw (R, C, W);} for (int i = 0, i < n; i++) for (int j = 0; J < N; j + +) if (g[i][j]! = ' * ') g[i][j] = '. '; for (int i = 0; i < n; i++) puts (G[i]);} int main () {//freopen ("T.txt", "R", stdin); int rnd = 0;while (~scanf ("%d", &n) && N) {memset (g, ' + ', sizeof (g)) CodE.clear (); memset (node, 0, sizeof (node)), if (RND) puts ("");p rintf ("Image%d\n", ++rnd), if (n > 0) {for (int i = 0; I < n; i++) scanf ("%s", G[i]), int id = 1;encode (0, 0, N, 0, id);//encode the entire image vector<int>tmp;for (int i = 1; i < ID; i++) if (Node[i].color = = 1) tmp.push_back (node[i].dir); sort (Tmp.begin (), Tmp.end ()); for (int i = 0; i < tmp.size (); i++) {print F ("%d", tmp[i]); if ((i + 1)% = = 0 | | i = = tmp.size ()-1) printf ("\ n"), Else printf (""); printf ("Total number of black nodes =%d\n", tmp.size ());} Else{n =-n;int X;while (~scanf ("%d", &x) && x! =-1) code.push_back (x);d Ecode (code);//decoding after drawing}}return 0;}

Exercise 6-8 Spatial Structure UVa806