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();
}