CERC2017 H Hidden Hierarchy(树+模拟)

题意：

在一些给定的目录里按要求展开到制定大小并按字典序输出

思路：

因为有目录这个东西，所以想到模拟一个类似字典树的东西，不过这里每个儿子可能有n个节点，而且不能O(1)查询了

代码超长。。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>
    
#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-6;
const int mod = 1e9+7;
const int maxn = 2e6+100;
const int maxm = 2e6+100;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);



//vector<string>str;//每个编号对应的字符串
string str[90000 + 10];
bool cmp(int a, int b){
    return str[a] < str[b];
}
vector<int>v[90000 + 10];//v[i]为i的孩子们
int tol = 1;
int vis[90000 + 10];//结尾节点
int vl[90000 + 10];//节点的值
int T;
void insert(vector<string>s, int val){
    int root = 0;
    int sz = s.size();
    for(int i = 0; i < sz; i++){
        //cout << i << "   "<<s[i]<<endl;
        int flg = 0;
        for(int j = 0; j < (int)v[root].size(); j++){
            if(str[v[root][j]] == s[i]){
                root = v[root][j];
                flg = 1;
                break;
            }
        }
        if(!flg){
            //cout << "  bug  "<<tol<<endl;
            v[root].pb(tol);
            str[tol] = s[i];
            if(i==(sz-1)){
                vis[tol] = 1;
                vl[tol] = val;
            }
            root = tol++;
        }
    }
    return;
}
int size[90000 + 10];
int readydfs(int root){
    int sz = v[root].size();
    int res = 0;
    if(sz == 0){
        return vl[root];
    }
    int flg = 0;
    //cout << root << "  "<<str[root]<<endl;
    for(int i = 0; i < sz; i++){
        //cout << v[root][i] << "  " << str[v[root][i]]<<endl;
        if(vis[v[root][i]] != 1){
            flg = 1;
        }
        res += readydfs(v[root][i]);
    }
    if(!flg) vis[root] = 2;

    return vl[root] = res;
}
void dfs(int root, string s, int c){
    int sz = v[root].size();
    int flg = 0;
    if(vis[root]==2)flg = 1;
    if(vl[root] < T){
        if(vis[root] == 2) flg = 1;
        else flg = 2;
    }
    //cout << s <<endl;
    /*if(flg == 0){
        cout<<"- "<<s<<"/ "<<vl[root]<<endl;
    }*/
    if(flg == 1){
        cout <<"  "<<s<<"/ "<<vl[root]<<endl;
        return;
    }
    else if(flg == 2){
        cout << "+ "<<s<<"/ "<<vl[root]<<endl;
        return;
    }
    sort(v[root].begin(), v[root].end(), cmp);
    int vv = 1;
    for(int i = 0; i < sz; i++){
        if(vis[v[root][i]]!=1){
            if(vv &&vl[v[root][i]]>=T){
                cout<<"- "<<s<<"/ "<<vl[root]<<endl;
                vv = 0;
                break;
            }
            //if(flg == 3) dfs(v[root][i], s+"/"+str[v[root][i]], 0);
            
        }
    }
    if(!vv){
        for(int i = 0; i < sz; i++){
            if(vis[v[root][i]]!=1){
            
                dfs(v[root][i], s+"/"+str[v[root][i]], 0);
            
            }
        }
    }
    else cout << "+ "<<s<<"/ "<<vl[root]<<endl;
    return;
}
int main() {
    tol = 1;
    str[0] = "/";
    mem(vis, 0);
    mem(vl, 0);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    string tmp;
    for(int t = 1; t <= n; t++){
        int val;
        cin >> tmp >> val;
        int len = tmp.size();
        vector<string>s;
        int p = 0;
        for(int i = 1; i < len; i++){
            if(i && (tmp[i] == '/')){
                s.pb(tmp.substr(p+1, i-p-1));
                p = i;
            }
            if(i == len - 1){
                s.pb(tmp.substr(p+1, i-p));
            }
        }
        insert(s, val);
    }
    readydfs(0);
    /*for(int i = 0; i < tol; i++){
        cout << i <<"  "<<vl[i]<<endl;
    }*/
    cin >> T;
    string s = "";
    dfs(0, s, 0);
    return 0;
}