A - 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
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
Input0010100Output
3Input
3 1 3 4
3 2 5 6
1 7
111Output
-1
蛇皮走位
構造一個方陣,對行來說 0就前進,1就後退
#include <bits/stdc++.h> using namespace std; typedef long long ll; #define inf 2147483647 const ll INF = 0x3f3f3f3f3f3f3f3fll; #define ri register int template <class T> inline T min(T a, T b, T c) { return min(min(a, b), c); } template <class T> inline T max(T a, T b, T c) { return max(max(a, b), c); } template <class T> inline T min(T a, T b, T c, T d) { return min(min(a, b), min(c, d)); } template <class T> inline T max(T a, T b, T c, T d) { return max(max(a, b), max(c, d)); } #define scanf1(x) scanf("%d", &x) #define scanf2(x, y) scanf("%d%d", &x, &y) #define scanf3(x, y, z) scanf("%d%d%d", &x, &y, &z) #define scanf4(x, y, z, X) scanf("%d%d%d%d", &x, &y, &z, &X) #define pi acos(-1) #define me(x, y) memset(x, y, sizeof(x)); #define For(i, a, b) for (int i = a; i <= b; i++) #define FFor(i, a, b) for (int i = a; i >= b; i--) #define bug printf("***********\n"); #define mp make_pair #define pb push_back const int N = 200005; // name******************************* vector<int>vec[N]; char s[N]; // function****************************** //*************************************** int main() { // ios::sync_with_stdio(0); // cin.tie(0); // freopen("test.txt", "r", stdin); // freopen("outout.txt","w",stdout); scanf("%s",s+1); int len=strlen(s+1); int r=0,k=0; For(i,1,len) { if(s[i]==‘0‘) vec[++r].pb(i); else { if(r<=0) { cout<<-1; return 0; } vec[r--].pb(i); } k=max(k,r); } if(k!=r) { cout<<-1; return 0; } cout<<k<<endl; For(i,1,k) { cout<<vec[i].size()<<" "; For(j,0,vec[i].size()-1) { cout<<vec[i][j]<<" "; } cout<<endl; } return 0; }
A - Zebras