| Time Limit: 3000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
Description
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:
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
题解：给你一个先序二叉树，其中左子结点在父节点左一个单位，右节点在父节点右一个单位；让从左到右输出每个水平位置的权值和；本来用的从1开始root<<1建树，发现思路行不通；因为每个节点的root不同，所以考虑到从MAXN/2开始，-1，+1进行建树不就可以了么，每次如果不等于-1，就加上权值，然后找到左边最小的root，右边最大的root输出答案即可；
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
using namespace std;
#define SI(x) scanf("%d",&x)
#define mem(x,y) memset(x,y,sizeof(x))
#define PI(x) printf("%d",x)
#define P_ printf(" ")
const int INF=0x3f3f3f3f;
typedef long long LL;
const int MAXN=100010;
int tree[MAXN];
int flot,cnt;
int l,r;
void solve(int root){
int x;
scanf("%d",&x);
if(!cnt&&x==-1){
flot=0;return;
}
cnt=1;
if(x!=-1){
l=min(l,root);
r=max(r,root);
tree[root]+=x;
solve(root-1);
solve(root+1);
} }
int main(){
int kase=0;
while(true){
flot=1;
memset(tree,0,sizeof(tree));
l=r=MAXN/2;
cnt=0;
solve(MAXN/2);
if(!flot)break;
printf("Case %d:\n",++kase);
for(int i=l;i<=r;i++){
if(i!=l)P_;printf("%d",tree[i]);
}
puts("\n");
}
return 0;
}