【codeforces 747E】Comments

【题目链接】:http://codeforces.com/problemset/problem/747/E

【题意】



给你一个类似递归的结构;

让你把每一层的字符串按照层,一层层地输出出来;

并输出层数;

【题解】



递归里面,写:

当前层开始的位置,这一层里面有多少个字符串,以及层的深度;

每次搞到下一层最右的下标就好;

然后每找到一个字符就加入到这一层里面;



【Number Of WA】



0



【完整代码】

#include <bits/stdc++.h>

using namespace std;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define rep1(i,a,b) for (int i = a;i <= b;i++)

#define rep2(i,a,b) for (int i = a;i >= b;i--)

#define mp make_pair

#define pb push_back

#define fi first

#define se second

#define ms(x,y) memset(x,y,sizeof x)

#define Open() freopen("F:\\rush.txt","r",stdin)

#define Close() ios::sync_with_stdio(0),cin.tie(0)

typedef pair<int, int> pii;

typedef pair<LL, LL> pll;

const int dx[9] = { 0,1,-1,0,0,-1,-1,1,1 };

const int dy[9] = { 0,0,0,-1,1,-1,1,-1,1 };

const double pi = acos(-1.0);

const int N = 1e6 + 100;

string temp;

char s[N];

int len, ma;

vector <string> ans[N];

void dfs(int i, int should, int dep)

{

    int j = i, num = 0;

    while (1)

    {

        if (j>len) return;

        temp = "";

        while (j <= len && s[j] != ',')

        {

            temp += s[j];

            j++;

        }

        ans[dep].pb(temp);

        LL temp1 = 0;

        j++;

        while (j <= len && s[j] != ',')

        {

            temp1 = temp1 * 10 + s[j] - '0';

            j++;

        }

        ma = max(ma, j + 1);

        if (temp1 == 0)

        {

            j++;

        }

        else

        {

            ma = j + 1;

            dfs(j + 1, temp1, dep + 1);

            j = ma;

        }

        num++;

        if (num == should) return;

    }

}

int main()

{

    //Open();

    Close();//scanf,puts,printf not use

            //init??????

    cin >> (s + 1);

    len = strlen(s + 1);

    dfs(1, len, 1);

    int dep = 1;

    while (int(ans[dep + 1].size())) dep++;

    cout << dep << endl;

    rep1(i, 1, dep)

    {

        int len = ans[i].size();

        rep1(j, 0, len - 1)

        {

            cout << ans[i][j];

            if (j == len - 1)

                cout << endl;

            else

                cout << ' ';

        }

    }

    return 0;

}