Atcoder wide flip解題報告

Atcoder wide flip

題目原文

You are given a string S consisting of 0 and 1.Find the maximum integer K not greater than |S| such that we can turn all the characters of S into 0 by repeating the following operation some number of times.

• Choose a contiguous segment [l,r] in S whose length is at least K (that is, r−l+1≥K must be satisfied). For each integer i such that l≤i≤r, do the following: if S_i is 0, replace it with 1; if S_i is 1, replace it with 0.

Constraints

• 1≤|S|≤10⁵
• S_i(1≤i≤N) is either 0 or 1.

題意分析

給定一個只有0,1兩個字符的字符串，每次翻轉其中[l,r]之間的位，且要滿足r-l+1>=K，求最大的K，使得字符串S最終被全部變爲0.

解法分析

其實全部變爲0或者1都一樣，因爲只需要全部再翻轉一次，翻轉次數肯定大於等於K，不影響K的最大值。此題也是動態規劃問題，假設下標爲i的位之前都已經相同，且S[i-1]!=S[i]，則爲了達到最終目的，一定會選擇翻轉前i個元素或者翻轉後n-i個元素，由於K要取最大值，因此取max(i,n-i)，遍歷一遍S，res取min(res,max(i,n-i))。C++代碼如下：

    #include<iostream>
    #include<string>
    #include<algorithm>
    using namespace std;
    int main(){
        string S;
        cin>>S;
        int res=S.size();
        int a=S.size();
        for(int i=1;i<S.size();i++){
            if(S[i]!=S[i-1])
                res=min(res,max(i,a-i));
        }
        cout<<res<<endl;
        return 0;
    }