题目链接：https://codeforces.com/problemset/problem/950/C
博客园食用链接：https://www.cnblogs.com/lonely-wind-/p/13453216.html

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 ∣ ) k (1 ≤ k ≤ |s|) 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 ∣ ) l _i (1 ≤ l _i ≤ |s|) li​(1 ≤ li​ ≤ ∣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.

Examples
Input
0010100
Output
3
3 1 3 4
3 2 5 6
1 7

Input
111
Output
-1

题目大意：给你一个序列，问你能否将其拆分为若干个斑马序列，若能请输出其拆分方法，否则输出-1，斑马序列的定义如下：开头是0，结尾是0，整个序列是01交替的，当然单个的0也是合法的。

emmm，不要想太多了，直接按照题目要求的来构造就行了，我们知道，1一定要放在0后面的，0也一定要放在1后面的（如果有1的话），然后多出来的0就单独成一段。

那么就可以用 v e c t o r < i n t > g [ m a x n ] vector<int>g[maxn] vector<int>g[maxn]来保存序列，假设现在的序列个数已经跟新到了第 c n t cnt cnt个，那么每次出现0的时候我们查询是否在 g [ 1 − c n t ] g[1-cnt] g[1−cnt]中是否有以1为结尾的，如果有我们就直接接上去，如果没有就单独开创，同时 c n t + + cnt++ cnt++，如果出现的是1，那么我们出现之前是否有以0为结尾的，如果有就接上去，没有就直接-1了。

现在有个问题就是怎么快速地查询这些以01为结尾的编号，如果for一遍的话时间复杂度有点高，所以我们可以用两个队列来保存这些编号，每次都将队首给弹出去就好了。

以下是AC代码：

#include <bits/stdc++.h>
using namespace std;

const int mac=2e5+10;

char s[mac];
vector<int>g[mac];
queue<int>tail1,tail0;

int main(int argc, char const *argv[])
{
	scanf ("%s",s);
	int len=strlen(s);
	int one=0,zero=0;
	for (int i=0; i<len; i++)
		if (s[i]=='0') zero++;
		else one++;
	if (one>=zero) {printf("-1\n"); return 0;}
	int cnt=0,mk=0;
	for (int i=0; i<len; i++){
		if (s[i]=='1' && tail0.empty()) {mk=1; break;}
		else if (s[i]=='1') {
			int v=tail0.front();
			tail0.pop();
			g[v].push_back(i+1);
			tail1.push(v);
		}
		else if (s[i]=='0'){
			if (tail1.empty()) {
				g[++cnt].push_back(i+1); 
				tail0.push(cnt);
				continue;
			}
			int v=tail1.front();
			tail1.pop();
			g[v].push_back(i+1);
			tail0.push(v);
		}
	}
	for (int i=1; i<=cnt; i++)
		if (s[g[i][g[i].size()-1]-1]=='1') {mk=1; break;}
	if (mk) {printf("-1\n"); return 0;};
	printf("%d\n",cnt);
	for (int i=1; i<=cnt; i++){
		printf("%d",g[i].size());
		for (auto x:g[i])
			printf(" %d",x);
		printf("\n");
	}
	return 0;
}