1006. Tree Traversals - Hard Version (35)解题报告

感谢神赐予我智慧和力量来解决这道题。

这道题对时间要求不高，用暴力DFS+适当的剪枝就可以解决。

中序LMR、前序MLR、后序LRM遍历序列有这样的性质，需要遍历的树上的任何一棵子树前序遍历序列的第一个元素等于后序遍历序列的最后一个元素，且该元素在中序遍历中恰好把根的左子树和右子树分割开来。

做题过程中参考了以下两篇博文：

PAT (Top Level) Practise 1006 Tree Traversals - Hard Version (35)

1006. Tree Traversals - Hard Version (35)

#define _CRT_SECURE_NO_WARNINGS#include <cstdio>#include <cstdlib>#include <cstring>#include <cctype>#include <queue>using namespace std;const int maxn = 110;int inorder[maxn], preorder[maxn], postorder[maxn], t[maxn], tree[maxn][2], f[maxn], root;int rein[maxn], repre[maxn], repost[maxn];bool flag;bool dfs(int &index, int inl, int inr, int prel, int prer, int postl, int postr);void dfs(int root, int p);void bfs(void);int get(void);int main(void) {int i, n, cnt = 0, residue;scanf("%d", &n);for (i = 1; i <= n; i++) inorder[i] = get(), f[inorder[i]]++, rein[inorder[i]] = i;for (i = 1; i <= n; i++) preorder[i] = get(), f[preorder[i]]++, repre[preorder[i]] = i;for (i = 1; i <= n; i++) postorder[i] = get(), f[postorder[i]]++, repost[postorder[i]] = i;for (i = 1; i <= n; i++) if (!f[i]) cnt++, residue = i;if(cnt > 1 || !dfs(root, 1, n, 1, n, 1, n)) puts("Impossible");else {for (i = 1; i <= n; i++) if (!t[i]) t[i] = residue;flag = 0, dfs(root, 2), putchar('\n');flag = 0, dfs(root, 1), putchar('\n');flag = 0, dfs(root, 3), putchar('\n');flag = 0, bfs();}return 0;}bool dfs(int &index, int inl, int inr, int prel, int prer, int postl, int postr) {int i, tmp;if (inl > inr) {index = 0; return true;}for (i = inl; i <= inr; i++) {if (inorder[i] && preorder[prel] && inorder[i] != preorder[prel]) continue;if (inorder[i] && postorder[postr] && inorder[i] != postorder[postr]) continue;if (preorder[prel] && postorder[postr] && preorder[prel] != postorder[postr]) return false;tmp = max(inorder[i], max(preorder[prel], postorder[postr]));if (rein[tmp] && rein[tmp] != i) continue;if (repre[tmp] && repre[tmp] != prel) continue;if (repost[tmp] && repost[tmp] != postr) continue;if (!dfs(tree[i][0], inl, i - 1, prel + 1, prel + i - inl, postl, postl + i - inl - 1)) continue;if (!dfs(tree[i][1], i + 1, inr, prel + i - inl + 1, prer, postl + i - inl, postr - 1)) continue;index = i;t[index] = tmp;return true;}return false;}void dfs(int root, int p) {if (!root) return;if (p == 1) printf("%s%d", flag ? " " : "", t[root]), flag = true;dfs(tree[root][0], p);if (p == 2) printf("%s%d", flag ? " " : "", t[root]), flag = true;dfs(tree[root][1], p);if (p == 3) printf("%s%d", flag ? " " : "", t[root]), flag = true;return;}void bfs(void) {queue<int> q;int tmp;q.push(root);printf("%d", t[root]);while (!q.empty()) {tmp = q.front();q.pop();if (tree[tmp][0]) q.push(tree[tmp][0]), printf(" %d", t[tree[tmp][0]]);if (tree[tmp][1]) q.push(tree[tmp][1]), printf(" %d", t[tree[tmp][1]]);}return;}int get(void) {int tmp, i;char str[10];scanf("%s", str);if (str[0] == '-') {return 0;}else {tmp = 0;for (i = 0; str[i]; i++) {tmp = tmp * 10 + str[i] - '0';}return tmp;}}