UVa 297 Quadtrees

Description

四分树是如果，全黑就f，如果全白就e，有黑有白就f 同时描述f的情况
描述顺序是
2 1
3 4
这样递归下去
现在给你两个四分树，把这两个四分树合并，最合并之后黑点的个数

Algorithm

直接根据四分树的定义把树建立在g数组上面
核心函数dfs就是建树过程
x,y表示左上顶点，ww表示方块宽度

本人版

#include <iostream>#include <cstring>using namespace std;const int maxn = 32 + 9;bool g[maxn][maxn];string s;int p;int ans;void dfs(int x, int y, int ww) {  char ch = s[p++];  if (ch == 'p') {    int w = ww / 2;    dfs(x,     y + w, w);    dfs(x,     y    , w);    dfs(x + w, y    , w);    dfs(x + w, y + w, w);    return;  }  if (ch == 'f') {    for (int i = x; i < x + ww; i++)      for (int j = y; j < y + ww; j++)        if (!g[i][j]) {          g[i][j] = true;          ans++;        }  }}void solve() {  memset(g, 0, sizeof(g));  p = 0;  ans = 0;  cin >> s;  dfs(0, 0, 32);  cin >> s;  p = 0;  dfs(0, 0, 32);  printf("There are %d black pixels.\n", ans);}int main() {  freopen("input", "r", stdin);  int t;  scanf("%d", &t);  for (int i = 0; i < t; i++) {    solve();  }}

佳佳版

#include <cstdio>#include <cstring>#include <iostream>using namespace std;const int len = 32;const int maxn = 1024;char s[maxn];int buf[len][len], cnt;void draw(const char* s, int &p, int r, int c, int w) {  char ch = s[p++];  if (ch == 'p') {    draw(s, p, r, c+w/2, w/2);    draw(s, p, r, c, w/2);    draw(s, p, r+w/2, c, w/2);    draw(s, p, r+w/2, c+w/2, w/2);  }else if (ch == 'f') {    for (int i = r; i < r+w; i++)      for (int j = c; j < c+w; j++)        if (buf[i][j] == 0) {          buf[i][j] = 1;          cnt++;        }  }}void solve() {  memset(buf, 0, sizeof(buf));  cnt = 0;  for (int i = 0; i < 2; i++) {    scanf("%s", s);    int p = 0;    draw(s, p, 0, 0, len);  }  printf("There are %d black pixels.\n", cnt);}int main() {  freopen("input","r",stdin);  int t;  scanf("%d", &t);  for (int i = 0; i < t; i++)    solve();}