Day6 - J - Cartesian Tree POJ - 2201
That is, if we denote left subtree of the node x by L(x), its right subtree by R(x) and its key by kx then for each node x we have
- if y ∈ L(x) then ky < kx
- if z ∈ R(x) then kz > kx
The binary search tree is called cartesian if its every node x in addition to the main key kx also has an auxiliary key that we will denote by ax, and for these keys the heap condition is satisfied, that is
- if y is the parent of x then ay < ax
Thus a cartesian tree is a binary rooted ordered tree, such that each of its nodes has a pair of two keys (k, a) and three conditions described are satisfied.
Given a set of pairs, construct a cartesian tree out of them, or detect that it is not possible.
Input
Output
The input ensure these is only one possible tree.
Sample Input
7
5 4
2 2
3 9
0 5
1 3
6 6
4 11
Sample Output
YES
2 3 6
0 5 1
1 0 7
5 0 0
2 4 0
1 0 0
3 0 0 思路:裸的笛卡尔树,学习新知识,这题输入唯一,一定有解,参考博客:https://blog.csdn.net/qq_36056315/article/details/79845193
https://blog.csdn.net/code92007/article/details/94591571
注意不要在退栈的时候改变fa指针就行,要根据退栈完毕后left和right指针进行fa的更改,不然已经定好的顺序会乱(
const int maxm = 5e4+; int fa[maxm], Left[maxm], Right[maxm], N; struct Node {
int key, value, id;
bool operator<(const Node &node) const {
return key < node.key;
}
} Nodes[maxm], s[maxm]; int main() {
scanf("%d", &N);
for(int i = ; i <= N; ++i) {
scanf("%d%d", &Nodes[i].key, &Nodes[i].value);
Nodes[i].id = i;
}
sort(Nodes+, Nodes+N+);
int top = ;
bool flag = false;
for(int i = ; i <= N; ++i) {
while(top && s[top].value > Nodes[i].value) {
Left[Nodes[i].id] = s[top].id;
top--;
}
fa[Nodes[i].id] = s[top].id;
fa[Left[Nodes[i].id]] = Nodes[i].id;
if(top)
Right[s[top].id] = Nodes[i].id;
s[++top] = Nodes[i]; }
printf("YES\n");
for(int i = ; i <= N; ++i)
printf("%d %d %d\n", fa[i], Left[i], Right[i]);
return ;
}
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