UVA 10410 Tree Reconstruction （树重建）

题意：给定一个树的bfs和dfs序列，升序输出每个结点的子结点列表。

分析：因为建树不唯一，假定若bfs[u] = bfs[v] + 1，则u是v的兄弟结点，否则是孩子结点。用栈维护。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 1000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int id[MAXN];
stack<int> s;
set<int> son[MAXN];
int main(){
    int n;
    while(scanf("%d", &n) == 1){
        while(!s.empty()) s.pop();
        memset(id, 0, sizeof id);
        for(int i = 0; i < MAXN; ++i) son[i].clear();
        int x;
        for(int i = 1; i <= n; ++i){
            scanf("%d", &x);
            id[x] = i;
        }
        int root;
        scanf("%d", &root);
        s.push(root);
        for(int i = 2; i <= n; ++i){
            scanf("%d", &x);
            while(1){
                int t = s.top();
                if(t == root || id[x] > id[t] + 1){
                    son[t].insert(x);
                    s.push(x);
                    break;
                }
                else s.pop();
            }
        }
        for(int i = 1; i <= n; ++i){
            printf("%d:", i);
            for(set<int>::iterator it = son[i].begin(); it != son[i].end(); ++it){
                cout << " " << *it;
            }
            printf("\n");
        }
    }
    return 0;
}