UVA - 10410 Tree Reconstruction

8
4 3 5 1 2 8 7 6
4 3 1 7 2 6 5 8

1: 7
2: 6
3: 1 2
4: 3 5
5: 8
6:
7:
8:

题意：告诉你DFS和BFS的结果，求树

思路：从DFS入手。每次枚举它之后的点有没有可能是它的子节点，推断从BFS推断，首先深度要是+1的，还有就是两个之间必须都要是訪问多的才行

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
const int MAXN = 1010;

int n, ord[MAXN], dfs[MAXN], bfs[MAXN], deep[MAXN];
vector<int> arr[MAXN];
queue<int> q;
int vis[MAXN];

int check(int a, int b, int curDeep) {
	int e = dfs[a];
	int f = dfs[b];
	int c = ord[e];
	int d = ord[f];
	for (int i = a+1; i < b; i++) {
		int tmp = dfs[i];
		if (deep[tmp] == 0)
			continue;
		if (deep[tmp] != curDeep)
			return 0;
	}
	for (int i = c+1; i < d; i++) {
		int tmp = bfs[i];
		if (!vis[tmp])
			return 0;
	}
	return 1;
}

void cal(int cur, int curDeep) {
	int cnt = dfs[cur];
	for (int i = cur+1; i < n; i++) {
		int tmp = dfs[i];
		if (vis[tmp])
			continue;
		if (check(cur, i, curDeep+1))
			arr[cnt].push_back(tmp);
		else continue;
		q.push(i);
		vis[tmp] = 1;
		deep[tmp] = curDeep + 1;
	}
}

void print() {
	for (int i = 1; i <= n; i++) {
		printf("%d:", i);
		for (int j = 0; j < arr[i].size(); j++)
			printf(" %d", arr[i][j]);
		printf("\n");
	}
}

int main() {
	while (scanf("%d", &n) != EOF) {
		memset(vis, 0, sizeof(vis));
		memset(deep, 0, sizeof(deep));
		memset(ord, 0, sizeof(ord));
		memset(dfs, 0, sizeof(dfs));
		memset(bfs, 0, sizeof(bfs));
		for (int i = 0; i <= n; i++)
			arr[i].clear();
		while (!q.empty())
			q.pop();
		for (int i = 0; i < n; i++) {
			int t;
			scanf("%d", &t);
			bfs[i] = t;
			ord[t] = i;
		}
		for (int i = 0; i < n; i++)
			scanf("%d", &dfs[i]);
		deep[dfs[0]] = 1;
		q.push(0);
		while (!q.empty()) {
			int cnt = q.front();
			q.pop();
			cal(cnt, deep[dfs[cnt]]);
		}
		print();
	}
	return 0;
}