CodeForces - 949A - Zebras
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
题目链接
这个题目就是给你一个0、1串，要求你找字串，可以是只有0，否则就需要是有前保0和最后也是0，如果中间有1，那么必须1之间有0间隔。至于找的子串只要符合要求怎么找都可以。
这个题目有点思维性，就是找子串的话，不一定要一次次的遍历找，可以申请一个vector数组，做一个动态二维数组，每一行行代表一个子串，每遇到一个0就把这个位置记录到当前行然后跳到下一行，如果遇到1，看是不是第一个数就是1，如果第一个就是的话，他一定没有前保0，直接输出-1，return就可以了，如果不是的话，就返回上一行，因为上一行的位置有前保0，然后记录该1的位置，这样把字符串遍历一遍就找完了。
AC代码
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int maxn = 2e5 + 5;
vector<int> v[maxn];
char str[maxn];
int main()
{
int cnt = 0;
gets(str);
int len = strlen(str);
int fuck = 0;
for(int i = 0; i < len; i++)
{
if(str[i] == '0') v[cnt++].push_back(i + 1);
else
{
if(!cnt)
{
printf("-1\n");
return 0;
}
v[--cnt].push_back(i + 1);
}
fuck = max(fuck, cnt);
}
if(fuck != cnt)
{
printf("-1\n");
return 0;
}
printf("%d\n", cnt);
for(int i = 0; i < cnt; i++)
{
cout << v[i].size();
for(int j = 0; j < v[i].size(); j++)
{
cout << " " << v[i][j];
}
printf("\n");
}
return 0;
}