简单点实现可以不用把各个节点连接起来建树,递归的扫描一遍,根据每个节点的表示的跨度范围和类型填充一个1024的数组,每个case中两个树依次填充完毕,最后统计数组中1的个数即为结果。
#include<stdio.h> char str[2000]; int num[1024]; int idx; void buildTree(char *s, int start, int end) { char type = s[idx++]; if (type == 'p') { int mid = (start + end) / 2; int lmid = (start + mid) / 2; int rmid = (mid + end) / 2; buildTree(s, start, lmid); buildTree(s, lmid + 1, mid); buildTree(s, mid + 1, rmid); buildTree(s, rmid + 1, end); } else if (type == 'f') { int i; for (i = start; i <= end; i++) num[i] = 1; } } int main() { int cases; scanf("%d", &cases); while (cases--) { idx = 0; int i; for (i = 0; i < 1024; i++) num[i] = 0; scanf("%s", str); buildTree(str, 0, 1023); idx = 0; scanf("%s", str); buildTree(str, 0, 1023); int sum = 0; for (i = 0; i < 1024; i++) { sum += num[i]; } printf("There are %d black pixels.\n", sum); } return 0; }