C. Zebras
512 megabytes
standard input
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
0010100
output
3
3 1 3 4
3 2 5 6
1 7
input
111
output
-1 题目大意
给定01串，需要把串中所有元素用上，且必须是"01"间隔或者"0"单独一个的形式
解题思路
模拟即可，用二维vector存取，每遇到0就放到最后一个元素为1的容器集合内；
若无，则创建一个新的容器集合；
如样例"0010100"
0 a[1][0]
0 a[2][0]
遇到1，则放a[2][1],接下来遇到0继续在a[2][2]补上,1 a[2][3]补上，0 a[2][4];
再遇到最后一个0，原理同第二个0，新建一个a[3][0]. 具体看代码 AC代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=2e5+;
char s[maxn];
vector<int>a[maxn];
int main()
{
scanf("%s",s+);
int len=strlen(s+);
int Max=-,ans=,flag=;
for(int i=;i<=len&&flag;i++)
{
if(s[i]=='') a[++ans].push_back(i);
else
{
if(ans==) flag=;
a[ans--].push_back(i);
//ans--;
}
Max=max(Max,ans);
}
if(Max!=ans) flag=;
if(flag==) puts("-1");
else
{
cout<<Max<<endl;
for(int i=;i<=Max;i++)
{
cout<<a[i].size();
for(int j=;j<a[i].size();j++)
printf(" %d",a[i][j]);
cout<<endl;
}
}
return ;
}