Each year, fall in the North Central region is accompanied by the brilliant colors of the leaves on the
trees, followed quickly by the falling leaves accumulating under the trees. If the same thing happened
to binary trees, how large would the piles of leaves become?
We assume each node in a binary tree ”drops” a number of leaves equal to the integer value stored
in that node. We also assume that these leaves drop vertically to the ground (thankfully, there’s no
wind to blow them around). Finally, we assume that the nodes are positioned horizontally in such a
manner that the left and right children of a node are exactly one unit to the left and one unit to the
right, respectively, of their parent. Consider the following tree on the right:
The nodes containing 5 and 6 have the same horizontal position
(with different vertical positions, of course). The node
containing 7 is one unit to the left of those containing 5 and
6, and the node containing 3 is one unit to their right. When
the ”leaves” drop from these nodes, three piles are created:
the leftmost one contains 7 leaves (from the leftmost node),
the next contains 11 (from the nodes containing 5 and 6), and
the rightmost pile contains 3. (While it is true that only leaf
nodes in a tree would logically have leaves, we ignore that in
this problem.)
Input
The input contains multiple test cases, each describing a single tree. A tree is specified by giving the
value in the root node, followed by the description of the left subtree, and then the description of the
right subtree. If a subtree is empty, the value ‘-1’ is supplied. Thus the tree shown above is specified
as ‘5 7 -1 6 -1 -1 3 -1 -1’. Each actual tree node contains a positive, non-zero value. The last test
case is followed by a single ‘-1’ (which would otherwise represent an empty tree).
Output
For each test case, display the case number (they are numbered sequentially, starting with 1) on a line
by itself. On the next line display the number of “leaves” in each pile, from left to right, with a single
space separating each value. This display must start in column 1, and will not exceed the width of an
80-character line. Follow the output for each case by a blank line. This format is illustrated in the
examples below.
Sample Input
5 7 -1 6 -1 -1 3 -1 -1
8 2 9 -1 -1 6 5 -1 -1 12 -1
-1 3 7 -1 -1 -1
-1
Sample Output
Case 1:
7 11 3
Case 2:
9 7 21 15
就是给你棵树，让你输出竖着的每一排节点的值的和
代码如下：
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 int sum[1000]; 5 void build (int pos) 6 { 7 int elem; 8 scanf("%d",&elem); 9 if (elem!=-1) 10 { 11 sum[pos]+=elem; 12 build(pos-1); 13 build(pos+1); 14 } 15 } 16 int main() 17 { 18 int root,casee=0; 19 bool f=false; 20 //freopen("de.txt","r",stdin); 21 while (~scanf("%d",&root)) 22 { 23 memset(sum,0,sizeof sum); 24 if (root==-1) 25 break; 26 f=true; 27 printf("Case %d: ",++casee); 28 sum[500]=root; 29 build(499); 30 build(501); 31 queue<int>q; 32 for (int i=0;i<1000;++i) 33 { 34 if(sum[i]!=0) 35 q.push(sum[i]); 36 } 37 bool f2=false; 38 while (!q.empty()) 39 { 40 if (f2) 41 printf(" "); 42 f2=true; 43 printf("%d",q.front()); 44 q.pop(); 45 } 46 printf(" "); 47 } 48 return 0; 49 }
各种PE23333333333