【链接】

【题意】

在这里输入题意

【题解】

发现根本不用存节点信息。

遇到了叶子节点且为黑色,就直接覆盖矩阵就好(因为是并集);

【代码】

#include 
    
     using namespace std;const int N = (1 << 5) + 10;string s;int pos, bo[N][N];void fugai(int x1, int y1, int x2, int y2) {    for (int i = x1; i <= x2; i++)        for (int j = y1; j <= y2; j++)            bo[i][j] = 1;}void dfs(int x1, int y1, int x2, int y2) {    if (s[pos] == 'f' || s[pos] == 'e') {        if (s[pos] == 'f') {            fugai(x1, y1, x2, y2);        }        return;    }    int len = (x2 - x1 + 1) / 2;    pos++;    dfs(x1, y1 + len, x1 + len - 1, y1 + len + len - 1);    pos++;    dfs(x1, y1, x1 + len - 1, y1 + len - 1);    pos++;    dfs(x1 + len, y1, x1 + len + len - 1, y1 + len - 1);    pos++;    dfs(x1 + len, y1 + len, x1 + len + len - 1, y1 + len + len - 1);}int main() {    //freopen("rush.txt", "r", stdin);    int T;    scanf("%d", &T);    while (T--) {        memset(bo, 0, sizeof bo);        cin >> s;        pos = 0;        dfs(1, 1, 32, 32);        cin >> s;        pos = 0;        dfs(1, 1, 32, 32);        int cnt = 0;        for (int i = 1; i <= 32; i++)            for (int j = 1; j <= 32; j++)                cnt += bo[i][j];        printf("There are %d black pixels.\n", cnt);    }    return 0;}