1143 Lowest Common Ancestor (30分)
程序员文章站
2022-07-14 14:49:27
...
在超时的边缘疯狂试探,注意有样例是u == v且存在于树中。
#include<vector>
#include<set>
#include<cstdio>
#define maxn 10010
using namespace std;
int pre[maxn];
struct node {
int data;
node *left, *right;
};
node* build_tree(int l, int r) {
if (l >= r) return NULL;
node* root = new node;
root->data = pre[l];
int i;
for (i = l + 1; i < r; i++) {
if (pre[i] >= pre[l]) break;
}
root->left = build_tree(l + 1, i);
root->right = build_tree(i, r);
return root;
}
bool finduv(node* root, int u, int v, bool &ue, bool &ve, int &a) {
if (root == NULL) return false;
bool lue = false, lve = false, rue = false, rve = false;
bool left = finduv(root->left, u, v, lue, lve, a);
if (left == true) return true;
bool right = finduv(root->right, u, v, rue, rve, a);
if (right == true) return true;
if (u == root->data || lue || rue) ue = true;
if (v == root->data || rve || lve) ve = true;
if (ue == true && ve == true) {
a = root->data;
return true;
}
return false;
}
int main() {
int M, N;
scanf("%d %d", &M, &N);
for (int i = 0; i < N; i++) {
scanf("%d", &pre[i]);
}
node* root = build_tree(0, N);
while (M--) {
int u, v;
scanf("%d %d", &u, &v);
bool ue = false, ve = false;
int a;
bool found = finduv(root, u, v, ue, ve, a);
if (found) {
if (a == u) {
printf("%d is an ancestor of %d.\n", u, v);
}
else if (a == v) {
printf("%d is an ancestor of %d.\n", v, u);
}
else printf("LCA of %d and %d is %d.\n", u, v, a);
}
else {
if (ue == false && ve == false) printf("ERROR: %d and %d are not found.\n", u, v);
else if (ue == false) printf("ERROR: %d is not found.\n", u);
else printf("ERROR: %d is not found.\n", v);
}
}
return 0;
}
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