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;
}