UVA 297 Quadtrees 好像不是很难。又好像不是很简单

UVA 297 Quadtrees

题目的意思挺好理解的，输入两颗四叉树，按照p为不空不满，f为满，e为空，按照先序输入， 把两颗四叉树合并起来，计算出新树的值。。

思路：首先把2个字符串处理，变成2个二维数组存放，然后在通过二维数组，就能查找到每个位置的结点相加之后的结果，要注意，如果四叉树4个分叉都为f，那么处理为结点为f，结点之后全部删除。

代码

#include <stdio.h>
#include <string.h>
#include <math.h>

int t;
char a[1366];
char b[1366];
int num;
int lena, lenb, lenc;
int l;
int la, lc;
char aa[6][1111];
char bb[6][1111];
char cc[6][1111];

void tra(char *a, char aa[][1111], int len, int x, int n, int m)
{
if (la == len)
{
return;
}
if (n % 4 == 0 && a[la - 1] != 'p' && la != 0)
{
tra(a, aa, len, x - 1, n / 4, m / 4);
}
if (a[la] == 'p')
{
aa[x]
= a[la];
la ++;
tra(a, aa, len, x + 1, n * 4, m * 4);
}
if (a[la] == 'f')
{
aa[x]
= a[la];
la ++;
tra(a, aa, len, x, n + 1 , m + 1);
}
if (a[la] == 'e')
{
aa[x]
= a[la];
la ++;
tra(a, aa, len, x, n + 1, m + 1);
}
}

int main()
{
scanf("%d", &t);
getchar();
while (t --)
{
l = 0;
num = 0;
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
memset(cc, 0, sizeof(cc));
memset(aa, 0, sizeof(aa));
memset(bb, 0, sizeof(bb));
gets(a);
gets(b);
lena = strlen(a);
lenb = strlen(b);
la = 0;
tra(a, aa, lena, 0, 0, 0);
la = 0;
tra(b, bb, lenb, 0, 0, 0);
for (int i = 5; i >= 0; i --)
for (int j = 0; j < pow(4, i); j ++)
{
if (i != 5)
{
if (cc[i + 1][j * 4] == 'f' && cc[i + 1][j * 4 + 1] == 'f' && cc[i + 1][j * 4 + 2] == 'f' && cc[i + 1][j * 4+ 3] == 'f')
{
cc[i][j] = 'f';
cc[i + 1][j * 4] = cc[i + 1][j * 4 + 1] = cc[i + 1][j * 4 + 2] = cc[i + 1][j * 4 + 3] = '\0';
continue;
}
}
if (aa[i][j] == 'f' || bb[i][j] == 'f')
{
cc[i][j] = 'f';
cc[i + 1][j * 4] = cc[i + 1][j * 4 + 1] = cc[i + 1][j * 4 + 2] = cc[i + 1][j * 4 + 3] = '\0';
continue;
}
if (aa[i][j] == 'e')
{
cc[i][j] = bb[i][j];
continue;
}
if (bb[i][j] == 'e')
{
cc[i][j] = aa[i][j];
continue;
}
if (aa[i][j] == 'p' && bb[i][j] == 'p')
{
cc[i][j] = 'p';
continue;
}
if (aa[i][j] == '\0' && bb[i][j] != '\0')
{
cc[i][j] = bb[i][j];
continue;
}
if (aa[i][j] != '\0' && bb[i][j] == '\0')
{
cc[i][j] = aa[i][j];
continue;
}
}
for (int i = 5; i >= 0; i --)
for (int j = 0; j < pow(4, i); j ++)
{
if (cc[i][j] == 'f')
num += (1024 / pow(4, i));
}
printf("There are %d black pixels.\n", num);
}
return 0;
}