Educational Codeforces Round 55 (Rated for Div. 2) B. Vova and Trophies 【贪心】

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探讨了在一次交换机会下，如何通过贪心算法找到最长连续'G'序列的解决方案，该问题源自Codeforces竞赛。

传送门：http://codeforces.com/contest/1082/problem/B

Vova has won

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.

Input

The first line contains one integer

The second line contains

Output

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

Examples

input

Copy

10
GGGSGGGSGG

output

Copy

input

Copy

4
GGGG

output

Copy

input

Copy

3
SSS

output

Copy

Note

In the first example Vova has to swap trophies with indices

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

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

题意概括：

给一串只含有 G 和 S 的字符串，有一次将两个字符对调位置的机会，求最长的连续 'G' 序列的长度。

解题思路：

想得太多系列，一开始用了二分搜索，debug到爆炸。

其实就是一个贪心：记录中间 ‘S’ 前面的 ‘G’的长度和后面 'G' 的长度，作和。取最大值 maxlen。最后答案和总的‘G’数量 snt 进行一下比较，如果大了说明多加的，输出 snt，否则输出 maxlen。

AC code：

  1 //#include <cstdio>
  2 //#include <iostream>
  3 //#include <cstring>
  4 //#include <algorithm>
  5 //#include <cmath>
  6 //#include <map>
  7 //#define INF 0x3f3f3f3f
  8 //#define LL long long
  9 //using namespace std;
 10 //const int MAXN = 1e5+10;
 11 //char str[MAXN];
 12 //int N, snt;
 13 //
 14 //bool check(int len)
 15 //{
 16 //    bool flag = false;
 17 //    int sum = 0, k = 0;
 18 //    int lst = 0;
 19 //    for(int i = 0; i < N; i++){
 20 //        if(str[i] == 'S' &&  i != 0 && str[i-1] == 'S'){
 21 //            sum = 0;k = 0;lst = 0;
 22 //        }
 23 //        if(str[i] == 'S' && !flag && str[i+1] == 'G' && i < N-1){
 24 //            flag = true;
 25 //            //printf("sum:%d\n", sum);
 26 //            lst = sum;
 27 //            if(sum <= snt)
 28 //                sum++;
 29 //            continue;
 30 //        }
 31 //        else if(str[i] == 'S' && flag == true && str[i-1] != 'S'){
 32 //            //printf("len:%d sum:%d\n",len, sum);
 33 //            if(sum >= len) return true;
 34 //            sum-=lst;
 35 //            lst=k;
 36 //            k=0;
 37 //        }
 38 //        else if(str[i] == 'G' && sum <= snt){
 39 //            //printf("len:%d i:%d sum:%d\n", len, i , sum);
 40 //            sum++;
 41 //            k++;
 42 //            if(sum >= len) return true;
 43 //        }
 44 //    }
 45 //    //printf("len:%d sum:%d\n", len, sum);
 46 //    //if(sum >= len) return true;
 47 //    return false;
 48 //}
 49 //
 50 //int main()
 51 //{
 52 //    scanf("%d", &N);
 53 //    scanf("%s", str);
 54 //    snt = 0;
 55 //    for(int i = 0; i < N; i++){
 56 //        if(str[i] == 'G') snt++;
 57 //    }
 58 //    //printf("%d\n", snt);
 59 //    int l = 0, r = N;
 60 //    int mid = 0, ans = 0;
 61 //    while(l<=r){
 62 //        //printf("mid:%d\n", mid);
 63 //        mid = (l+r)>>1;
 64 //        if(check(mid)) {
 65 //                l = mid+1;
 66 //                ans=mid;
 67 //        }
 68 //        else r = mid-1;
 69 //    }
 70 //    //printf("l:%d snt:%d\n", l, snt);
 71 //    if(ans == snt+1) ans = snt;
 72 //    printf("%d\n", ans);
 73 //    return 0;
 74 //}
 75 
 76 
 77 #include <cstdio>
 78 #include <iostream>
 79 #include <algorithm>
 80 #include <cmath>
 81 #include <cstring>
 82 #define INF 0x3f3f3f3f
 83 #define LL long long
 84 using namespace std;
 85 const int MAXN = 1e5+10;
 86 char str[MAXN];
 87 int N;
 88 int main()
 89 {
 90     scanf("%d", &N);
 91     scanf("%s", str);
 92     int gg = 0;
 93     int g = 0, k = 0;
 94     int len = 0;
 95 
 96     for(int i = 0; i < N; i++){
 97         if(str[i] == 'G'){
 98             g++;
 99             k++;
100         }
101         else{
102             gg = g;
103             g = 0;
104         }
105         len = max(len, gg+g+1);
106     }
107     //printf("%d %d\n", len, k);
108     int ans = min(len, k);
109     printf("%d\n", ans);
110     return 0;
111 }