CodeForces - 950C Zebras VirtualJudge 500题AC留念

Posted 2020-10-26 晓风微微

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 l_i (1 ≤ l_i ≤ |s|), which is the length of the i-th subsequence, and then l_i 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.

Example

Input

0010100

Output

3
3 1 3 4
3 2 5 6
1 7

Input

111

Output

-1

超级大模拟
碰到0，优先加入1结尾的数列，不然新建一个数列
碰到1，加入以0结尾的数列，如果没有以0结尾的数列，那么就不存在可xing的分配情况
最后再检测是否全部数列都以0结尾，如果存在以1结尾的数列那么不存在可行的分配情况
另外需要注意的是，建立一个1结尾队列索引和0结尾队列索引的数列，不需要每次去真的
找一遍0结尾和1结尾的队列

代码:

 1 #include<iostream>
 2 using namespace std;
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<vector>
 6 #include<cstdlib>
 7 const int maxn = 1e5*5;
 8 char s[maxn];
 9 vector<int> a[maxn];
10 vector<int> v0;
11 vector<int> v1;
12 int num = 0;
13 int main(){
14     scanf("%s",s);
15     int ok=1;
16     int len =strlen(s);
17     int pt;
18     for(int i=0;i<len;i++){
19         s[i] = s[i] - ‘0‘;
20         if(s[i]==0){
21             if(v1.size()==0){
22                 v0.push_back(num);
23                 a[num].push_back(i+1);
24                 num+=1;
25             }
26             else{
27                 pt = v1[v1.size()-1];
28                 v1.pop_back();
29                 v0.push_back(pt);
30                 a[pt].push_back(i+1);
31             }
32         }
33         else if(s[i]==1){
34             if(v0.size()==0){
35                 ok=0;
36                 break;
37             }
38             else{
39                 int pt = v0[v0.size()-1];
40                 v0.pop_back();
41                 v1.push_back(pt);
42                 a[pt].push_back(i+1);
43             }
44         }
45     }
46     if(v1.size()!=0)
47         ok = 0;
48     if(ok==0){
49         printf("-1");
50     }
51     else{
52         printf("%d\n",num);
53         for(int i=0;i<num;i++){
54             printf("%d",a[i].size());
55             for(int j=0;j<a[i].size();j++){
56                 printf(" %d",a[i][j]);
57             }
58             putchar(‘\n‘);
59         }
60     }
61     return 0;
62 }