PAT (Top Level) Practice1006 Tree Traversals - Hard Version (35 分)
阿新 • • 發佈:2018-12-23
題目連結:https://pintia.cn/problem-sets/994805148990160896/problems/994805154321121280
dfs加剪枝,很麻煩,具體看程式碼。
演算法:dfs
#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; } }