E. Comments dfs模拟

http://codeforces.com/contest/747/problem/E

首先，把字符串变成这个样子。

hello,2,ok,0,bye,0,test,0,one,1,two,2,a,0,b,0

hello 2 1
ok 0 2
bye 0 3
test 0 4
one 1 5
two 2 6
a 0 7
b 0 8

然后就可以按着每一位来dfs，第二个参数表示他有2个儿子，dfs的时候开个string ans[maxn]来保存每一层的字符串就可以了，

比赛的时候，傻傻逼逼地dfs构图，建好图后，又bfs，然后发现是头插法，又要reveser，很麻烦。不过幸好过了。

赛后补上这个方法。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#define root 1
const int maxn = 1e6 + 20;
struct node {
    string s;
    int ti;
    int id;
}a[maxn];
int lena;
char str[maxn];
string out[maxn];
string ans[maxn];
int up;
int mxdeep;
void dfs(int cur, int deep) {
    up = max(up, cur);
    mxdeep = max(mxdeep, deep);
    ans[deep] += out[cur];
    ans[deep] += " ";
    for (int i = 1; i <= a[cur].ti; ++i) {
        dfs(up + 1, deep + 1);
    }
}
void work() {
    scanf("%s", str + 1);
    int lenstr = strlen(str + 1);
    int now = 0;
    string ts;
    int di = 0;
    int to = 0;
    lenstr += 1;
    str[lenstr] = ',';
    lena = 0;
    for (int i = 1; i <= lenstr; ++i) {
        if (now == 0) {
            if (str[i] == ',') {
                now = 1;
            } else ts += str[i];
        } else {
            if (str[i] == ',') {
                ++lena;
                a[lena].s = ts;
                a[lena].ti = di;
                a[lena].id = ++to;
                out[to] = ts;
                di = 0;
                ts.clear();
                now = 0;
            } else di = di * 10 + str[i] - '0';
        }
    }
//    for (int i = 1; i <= lena; ++i) {
////        printf("%d %d %d")
//        cout << a[i].s << " " << a[i].ti << " " << a[i].id << endl;
//    }
    up = 1;
    while (up <= lena) {
        dfs(up, 1);
        up++;
    }
    printf("%d
", mxdeep);
    for (int i = 1; i <= mxdeep; ++i) {
        cout << ans[i] << endl;
    }
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    work();
    return 0;
}