[gym102267L]ABC

dxyinme

于 2019-10-03 10:10:40 发布

阅读量424

点赞数

CC 4.0 BY-SA版权

分类专栏： -----------------------归纳

本文链接：https://blog.youkuaiyun.com/dxyinme/article/details/101977313

-----------------------归纳专栏收录该内容

13 篇文章

订阅专栏

本文介绍了一个字符串操作问题，目标是通过特定操作将给定的字符串变为空串。文章详细阐述了解决方案，包括使用栈来维护字符串状态，并给出了具体实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 $" a b "$ .
Replace a character $^{'} b^{'}$ in the string with $" b c "$ .
Replace a character $^{'} c^{'}$ in the string with $" b a "$ .
Remove a substring(consecutive characters) $" a b c "$ from the string.

Let $n$ be the length of the string, can you remove the whole string using at most $3 n$ 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 $m (1 \leq m \leq 3 n)$ , 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

Input

bac

Output

-1

Note

This is how the string changes in the first example: $a c a b \to a b c a b \to a b \to a b c \to ϕ$ , where $ϕ$ is the empty string.

题意：
给定一个字符串，只含 $^{'} a^{'},^{'} b^{'},^{'} c^{'}$ 三种字符
你有四种操作
1.将一个’a’,变成"ab"
2.将一个’b’,变成"bc"
3.将一个’c’,变成"ba"
4.删除一个"abc"子串
设字符串长度为n，则你的操作数不能超过3*n
输出如何操作才能将给定的字符串变为空串
格式为 $type_i \ \ index_i$ 分别表示第i次操作的操作类型和操作位置。
如果不能则输出-1

题解:
考虑用栈来维护当前字符串。

首先我们考虑删去 $c$
考虑结尾的情况
对于ac来说，我们可以 $a c \to a b a \to a b c a \to a$
读于bbc来说，我们可以 $b b c \to b c b c \to b b a b c \to b b$
对于abc来说，我们直接删掉abc
对于bc来说，我们无法删去c，直接-1

然后处理完所有的c
我们得到了一个ab串
然后对于每个b来说只要找一个a对应即可。

最后对于只剩a的情况，直接删掉即可, $a \to a b \to a b c \to ϕ$

#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;
}