Vova and Trophies

题意翻译

题目描述

你现在有 nn 枚奖牌，每枚奖牌为金牌或银牌。这些奖牌现在按顺序排成一排。现在你可以调换任意一对奖牌的位置，求金牌最长连续段。

输入格式

第一行一个整数 nn (2\le n\le 10^5)(2≤n≤105) .

第二行一行字符串，表示你的奖牌。 ( \texttt{G}G 表示金牌，\texttt{S}S 表示银牌)

输出格式

输出一个整数表示答案。

题目描述

Vova has won nn trophies in different competitions. Each trophy is either golden or silver. The trophies are arranged in a row.

The beauty of the arrangement is the length of the longest subsegment consisting of golden trophies. Vova wants to swap two trophies (not necessarily adjacent ones) to make the arrangement as beautiful as possible — that means, to maximize the length of the longest such subsegment.

Help Vova! Tell him the maximum possible beauty of the arrangement if he is allowed to do at most one swap.

输入输出格式

输入格式：

The first line contains one integer nn ( 2 \le n \le 10^52≤n≤105 ) — the number of trophies.

The second line contains nn characters, each of them is either G or S. If the ii -th character is G, then the ii -th trophy is a golden one, otherwise it's a silver trophy.

输出格式：

Print the maximum possible length of a subsegment of golden trophies, if Vova is allowed to do at most one swap.

输入输出样例

输入样例#1： 复制

10
GGGSGGGSGG

输出样例#1： 复制

输入样例#2： 复制

4
GGGG

输出样例#2： 复制

输入样例#3： 复制

3
SSS

输出样例#3： 复制

说明

In the first example Vova has to swap trophies with indices 44 and 1010 . Thus he will obtain the sequence "GGGGGGGSGS", the length of the longest subsegment of golden trophies is 77 .

In the second example Vova can make no swaps at all. The length of the longest subsegment of golden trophies in the sequence is 44 .

In the third example Vova cannot do anything to make the length of the longest subsegment of golden trophies in the sequence greater than 00 .

思路：记录字符串中每个s前面有多少个g，如果没有s，就只统计g的个数，然后保留到二维数组里，剩下的就是分情况讨论，1：如果某个s前后的g的个数总和小于总共g的个数，说明可以从其他地方拿一个，这样的话就可以将s前后的两个g串合起来然后长度+1；2：如果某个s前后的g的个数总和等于总共g的个数（比如ggggsggg）那么就是将一个串中拿出一个g替换s，这样长度就是两个g串长度的和；3，这个就比较简单了，说明没有s，只有g，直接打印

#include<iostream>
#include<queue>
#include<string.h>
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long int LL;
const int inf = 0x7f7f7f7f;
const int N=2e5+5;
int num[N];
int main()
{int n;vector<int>ve;int cnt=0;cin>>n;string str;cin>>str;int sumg=0;int count=0;for(int i=0;i<n;i++){if(str[i]=='S'){//遇到s，就将之前的g的个数保存的数组里ve.push_back(count);count=0;}else{count++;//用来统计每个s前面g的个数}}for(int i=0;i<n;i++){//记录g的总共个数if(str[i]=='G')sumg++;}if(str[n-1]=='G'){//特判一下最后一个不是Sve.push_back(count);}int ans=0;if(ve.size()==1){//cout<<ve[0]<<endl;return 0;}else{for(int i=0;i<ve.size()-1;i++){if(ve[i]+ve[i+1]<sumg)//第一种情况ans=max(ans,ve[i]+ve[i+1]+1);else{ans=max(ans,ve[i]+ve[i+1]);//第二种情况}}
}
cout<<ans<<endl;return 0;}