[gym102267L]ABC

time limit per test : 1.0 s
memory limit per test : 256 MB

You are given a string consisting of letters $'a', 'b'$ and $'c'$, and there are $4$ kinds of operations you can do on it:

Replace a character $'a'$ in the string with $"ab"$.
Replace a character $'b'$ in the string with $"bc"$.
Replace a character $'c'$ in the string with $"ba"$.
Remove a substring(consecutive characters) $"abc"$ from the string.

Let $n$ be the length of the string, can you remove the whole string using at most $3n$ operations or state that it’s impossible to do so?
Input

The first and only line contains the string $s(1≤n≤2×10^5)$ consisting of characters $'a', 'b'$ and $'c'$.
Output

If it’s impossible to remove the whole string print $-1$, otherwise in the first line print $m(1≤m≤3n)$, the number of operations you will make.

In each of the next $m$ lines print an operation of the form $type_i,index_i(1≤type_i≤4,1≤index_i≤|s|)$, the type of the ith operation and the index of the character you want to do the ith operation on, if the operation is of type $4$, then indexi should be the index of the first character of the substring “abc” that you want to remove. Indexi is $1$−based and the string is updated after each operation, see example notes for better understanding.
Examples
Input

acab

Output

4
1 1
4 1
2 2
4 1

Input

bac

Output

-1

Note

This is how the string changes in the first example: $acab→abcab→ab→abc→ϕ$, where $ϕ$ is the empty string.

題意：
給定一個字符串，只含$'a','b','c'$三種字符
你有四種操作
1.將一個’a’,變成"ab"
2.將一個’b’,變成"bc"
3.將一個’c’,變成"ba"
4.刪除一個"abc"子串
設字符串長度爲n，則你的操作數不能超過3*n
輸出如何操作才能將給定的字符串變爲空串
格式爲$type_i \ \ index_i$分別表示第i次操作的操作類型和操作位置。
如果不能則輸出-1

題解:
考慮用棧來維護當前字符串。

首先我們考慮刪去$c$
考慮結尾的情況
對於ac來說，我們可以$ac→aba→abca→a$
讀於bbc來說，我們可以$bbc→bcbc→bbabc→bb$
對於abc來說，我們直接刪掉abc
對於bc來說，我們無法刪去c，直接-1

然後處理完所有的c
我們得到了一個ab串
然後對於每個b來說只要找一個a對應即可。

最後對於只剩a的情況，直接刪掉即可, $a→ab→abc→ϕ$

#include<bits/stdc++.h>
#define ll long long
#define pa pair<int,int>
using namespace std;
char s[200004];
char st[200004],st2[200004];
vector<pa>ans;
int l,t,t2;
int main(){
    scanf("%s",s+1);
    l=strlen(s+1);
    t=0;
    if(s[1]!='a'){
        return puts("-1"),0;
    }
    for(int i=1;i<=l;i++){
        if(s[i]=='c'){
            if(t>1){
                if(st[t-1]=='a'&&st[t]=='b'){
                    ans.push_back({4,t-1});
                    t-=2;
                    continue;
                }
                if(st[t]=='a'){
                    ans.push_back({3,t+1});
                    ans.push_back({2,t+1});
                    ans.push_back({4,t});
                    continue;
                }
                if(st[t-1]=='b'&&st[t]=='b'){
                    ans.push_back({2,t-1});
                    ans.push_back({3,t});
                    ans.push_back({4,t+1});
                    continue;
                }
                return puts("-1"),0;
            }
            else if(t==0){
                return puts("-1"),0;
            }
            else if(t==1){
                if(st[t]=='b'){
                    return puts("-1"),0;
                }
                else{
                    ans.push_back({3,t+1});
                    ans.push_back({2,t+1});
                    ans.push_back({4,t});
                    continue;
                }
            }
        }
        else{
            st[++t]=s[i];
        }
    }
    //for(int i=1;i<=t;i++)cout<<st[i];cout<<endl;
    t2=0;
    for(int i=1;i<=t;i++){
        if(st[i]=='b'){
            if(t2==0){
                return puts("-1"),0;
            }
            else{
                ans.push_back({2,t2+1});
                ans.push_back({4,t2});
                t2--;
            }
        }
        else{
            st2[++t2]=st[i];
        }
    }
    for(int i=1;i<=t2;i++){
        ans.push_back({1,1});
        ans.push_back({2,2});
        ans.push_back({4,1});
    }
    if(ans.size()>l*3)return puts("-1"),0;
    printf("%d\n",ans.size());
    for(int i=0;i<ans.size();i++)printf("%d %d\n",ans[i].first,ans[i].second);
    return 0;
}