uva 10729 - Treequivalence(暴力+离散）

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题目链接：uva 10729 - Treequivalence

题目大意：给出两棵树，判断两棵树是否相同，注意节点的字母可能相同，孩子的顺序不能变化，并且是多叉树。

解题思路：这题还是比较恶心的，因为节点的字母可能相同，所以根不能通过说记录节点的情况或者是求出前中序来判断两棵树是否相同。只能暴力枚举，对于每种节点用map离散记录，并且有一个特定的key，为遍历的顺序（唯一的，可以不为遍历顺序）。

#include <stdio.h>#include <string.h>#include <algorithm>#include <vector>#include <map>using namespace std;const int N = 150;struct Tree {int cnt;char node[N];vector<int> g[N];void clear() {cnt = 0;memset(node, 0, sizeof(node));for (int i = 0; i < N; i++) g[i].clear();}};int read(Tree& T, int far) {int x = T.cnt++;if (far >= 0) T.g[x].push_back(far);char ch;scanf(" %c %c", &T.node[x], &ch);if (ch != '(') {ungetc(ch, stdin);return x;}while (true) {T.g[x].push_back(read(T, x));if (scanf(" %c", &ch) != 1 || ch != ',') break;}return x;}void readTree(Tree& T) {T.clear();read(T, -1);}map<vector<int>, int> rec;int solve (const Tree& t, int x, int far = -1) {vector<int> v;if (far < 0) {for (int i = 0; i < t.g[x].size(); i++) {int u = t.g[x][i];v.push_back(solve(t, u, x));}int p = 0, n = v.size();for (int s = 1; s < n; s++) {for (int i = p, j = s, k = 0; k < n; k++) {if (v[i] != v[j]) {if (v[j] < v[i]) p = s;break;}if (++i >= n) i = 0;if (++j >= n) j = 0;}}rotate(v.begin(), v.begin() + p, v.end());} else {int i;for (i = 0; t.g[x][i] != far; i++);while (true) {if (++i >= t.g[x].size()) i = 0;int u = t.g[x][i];if (u == far) break;v.push_back(solve(t, u, x));}}v.push_back(t.node[x]);int key;if (rec.count(v) > 0) {key = rec[v];} else {key = rec.size();rec[v] = key;}return key;}bool judge(const Tree& T1, const Tree& T2) {if (T1.cnt != T2.cnt) return false;rec.clear();for (int i = 0; i < T1.cnt; i++) {if (solve(T1, i, -1) == solve(T2, 0, -1)) return true;}return false;}int main () {int cas;Tree T1, T2;scanf("%d", &cas);while (cas--) {readTree(T1);readTree(T2);printf("%s\n", judge(T1, T2) ? "same" : "different");}return 0;}

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