Codeforces Round #469 (Div. 2) C. Zebras

C. Zebras

time limit per test

1 second

memory limit per test

512 megabytes

input

standard input

output

standard output

Oleg writes down the history of the days he lived. For each day he decides if it was good or bad. Oleg calls a non-empty sequence of days a zebra, if it starts with a bad day, ends with a bad day, and good and bad days are alternating in it. Let us denote bad days as 0 and good days as 1. Then, for example, sequences of days 0, 010, 01010 are zebras, while sequences 1, 0110, 0101 are not.

Oleg tells you the story of days he lived in chronological order in form of string consisting of 0 and 1. Now you are interested if it is possible to divide Oleg's life history into several subsequences, each of which is a zebra, and the way it can be done. Each day must belong to exactly one of the subsequences. For each of the subsequences, days forming it must be ordered chronologically. Note that subsequence does not have to be a group of consecutive days.

Input

In the only line of input data there is a non-empty string s consisting of characters 0 and 1, which describes the history of Oleg's life. Its length (denoted as |s|) does not exceed 200 000 characters.

Output

If there is a way to divide history into zebra subsequences, in the first line of output you should print an integer k (1 ≤ k ≤ |s|), the resulting number of subsequences. In the i-th of following k lines first print the integer li (1 ≤ li ≤ |s|), which is the length of the i-th subsequence, and then li indices of days forming the subsequence. Indices must follow in ascending order. Days are numbered starting from 1. Each index from 1 to n must belong to exactly one subsequence. If there is no way to divide day history into zebra subsequences, print -1.

Subsequences may be printed in any order. If there are several solutions, you may print any of them. You do not have to minimize nor maximize the value of k.

Examples

input

Copy

0010100

output

3
3 1 3 4
3 2 5 6
1 7

input

Copy

111

output

-1

题意其实很简单，要求从给的01组成的字符串中找出以0开头和结束，中间01交错的子串，并返回符合条件的子串长度以及下标，下标要求升序。并且由于有多解，所以输出符合条件的解就可以。（这句话让我把我好不容易改对的代码因为跟样例形式不一样又多改了半小时）。
我的思路就是利用vector可以查看大小的方便性质（其实开个数组给数据计数也行，就是麻烦），把0和1交替存入vector，存的方式可以理解成，出现0，计数器cnt++，出现1，计数器cnt--，这样就保证了0和1可以交错出现，每次push_back操作的时候维护一个cnt的最大值，它就是开出来的vector的数量，而由于0开头0结束，那么cnt最后经过加减之后的大小就是有效的vector数量，然后输出就可以了。
还有一个坑点就是，v.size()这个操作其实是很耗时间的，一定要先把值存下来，在进入后面的运算，（我因为这个tle了一次）。

#include<bits/stdc++.h>

using namespace std;
vector<int>ans[200005];
char s[200005];
int cnt=0,sum=0;

int main()
{
    scanf("%s",s);
    int l=strlen(s);
    for(int i=0; i<l; i++)            //注意下标要返回0-1，所以push_back的时候要+1
    {
        if(s[i]=='0')
        {
            ans[cnt].push_back(i+1);
            cnt++;
        }
        else
        {
            if(cnt==0)
            {
                printf("-1\n");
                return 0;
            }
            cnt--;
            ans[cnt].push_back(i+1);
        }
        sum=max(cnt,sum);
    }
    if(cnt!=sum)
        printf("-1\n");
    else
    {
        printf("%d\n",sum);
        for(int i=0; i<sum; i++)
        {
            int len=ans[i].size();
            printf("%d ",len);
            for(int j=0; j<len; j++)
            {
                printf("%d",ans[i][j]);
                if(j<len-1)    printf(" ");
            }
            printf("\n");
        }
    }
    return 0;
}