问题 I: Stones

2024-01-26 07:48

文章标签 问题 stones

本文主要是介绍问题 I: Stones，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

题目描述

There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is ., and the stone is black if the character is #.

Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.

Constraints
1≤N≤2×105
S is a string of length
N consisting of . and #.

输入

Input is given from Standard Input in the following format:

N
S

输出

Print the minimum number of stones that needs to be recolored.

样例输入
复制样例数据 3
#.#

样例输出
1

思路：被这题卡了一个小时，是一开始的思路不对，想得太复杂了，还分类讨论，其实只需要暴力算法就可以通过了，就是将一块石头的左边全部变为white，右边全部为black，然后更新一下答案即可

#include <iostream>
#include <string>
#include <map>
#include <cstdio>
#include <algorithm>
using namespace std;char a[200005];
int black[200005],beginn,endd;
int cntb=1;
int main()
{int n,enddd;cin>>n;getchar();int ans=n;for(int i=1;i<=n;i++) {scanf("%c",&a[i]);if(a[i]=='#') black[cntb++]=i;}cntb--,beginn=black[1],endd=black[cntb];//cout<<cntb<<endl;int CNT=0;for(int i=1;i<=cntb;i++){int cnt=i-1;int Cnt=0;int j;for(j=black[i];a[j]=='#';j++) Cnt++;i+=Cnt-1;CNT+=Cnt;int cntw=(n-j+1)-(cntb-CNT);//cout<<cntw<<endl;ans=min(cntw+cnt,ans);}ans=min(cntb,ans);cout<<ans<<endl;
}