You have a necklace of N red, white, or blue beads (3<=N<=350) some of which are red, others blue, and others white, arranged at random. Here are two examples for n=29:

                1 2                               1 2
            r b b r                           b r r b
          r         b                       b         b
         r           r                     b           r
        r             r                   w             r
       b               r                 w               w
      b                 b               r                 r
      b                 b               b                 b
      b                 b               r                 b
       r               r                 b               r
        b             r                   r             r
         b           r                     r           r
           r       r                         r       b
             r b r                             r r w
            Figure A                         Figure B
                        r red bead
                        b blue bead
                        w white bead

The beads considered first and second in the text that follows have been marked in the picture.

The configuration in Figure A may be represented as a string of b‘s and r‘s, where b represents a blue bead and r represents a red one, as follows: brbrrrbbbrrrrrbrrbbrbbbbrrrrb .

Suppose you are to break the necklace at some point, lay it out straight, and then collect beads of the same color from one end until you reach a bead of a different color, and do the same for the other end (which might not be of the same color as the beads collected before this).

Determine the point where the necklace should be broken so that the most number of beads can be collected.

Example

For example, for the necklace in Figure A, 8 beads can be collected, with the breaking point either between bead 9 and bead 10 or else between bead 24 and bead 25.

In some necklaces, white beads had been included as shown in Figure B above. When collecting beads, a white bead that is encountered may be treated as either red or blue and then painted with the desired color

Write a program to determine the largest number of beads that can be collected from a supplied necklace.

PROGRAM NAME: beads

INPUT FORMAT

SAMPLE INPUT (file beads.in)

29
wwwbbrwrbrbrrbrbrwrwwrbwrwrrb

OUTPUT FORMAT

A single line containing the maximum of number of beads that can be collected from the supplied necklace.

SAMPLE OUTPUT (file beads.out)

11

OUTPUT EXPLANATION

Consider two copies of the beads (kind of like being able to runaround the ends). The string of 11 is marked.

                Two necklace copies joined here
                             v
wwwbbrwrbrbrrbrbrwrwwrbwrwrrb|wwwbbrwrbrbrrbrbrwrwwrbwrwrrb
                       ******|*****
                       rrrrrb|bbbbb  <-- assignments
                   5xr .....#|#####  6xb

                        5+6 = 11 total

就是暴力，然而開始的時候有很多細節沒註意到。比如起始點如果是w，比如當flag>n時的處理……總之WA了好幾遍才A。而且覺得代碼寫的很復雜。

 1 #include <iostream>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <cstdio>
 5 #include <cstdlib>
 6 #include <algorithm>
 7 using namespace std;
 8 int a[355];
 9 int main()
10 {
11     freopen("beads.in","r",stdin);
12     freopen("beads.out","w",stdout);
13     int n;
14     scanf("%d",&n);
15     for(int i=1;i<=n;i++)
16     {
17         char s0;
18         cin>>s0;
19         //if(s0==‘ ‘) {i--;continue;}
20         if(s0==‘b‘) a[i]=2;
21         else if(s0==‘r‘) a[i]=1;  //這裏如果不是 else if而是 if就會錯 
22         else a[i]=0;
23     }
24     int maxn=0;
25     for(int i=1;i<=n;i++)
26     {
27         int flag=i,color=a[i];
28         int cnt=0;
29         if(color==0)   //起點為白 
30         {
31             int j=flag,c=0;
32             while(a[j]==0)
33             {
34                 j--;
35                 if(j<1) j=n;
36                 c++;
37                 if(c==n) {printf("%d\n",n);return 0;}  //全部為白 
38             }
39             color=a[j];
40         }
41         while(a[flag]==color || a[flag]==0)  //往前找 
42         {
43             cnt++;
44             flag--;               
45             if(flag<1) flag=n;
46             if(cnt==n) break;
47         }
48         flag=i+1;
49         if(i+1>n) {color=a[1];flag=1;}
50         else color=a[i+1];
51         if(color==0)
52         {
53             int j=flag,c=0;
54             while(a[j]==0)
55             {
56                 j++;
57                 if(j>n) j=1;
58                 c++;
59                 if(c==n) {printf("%d\n",n);return 0;}
60             }
61             color=a[j];
62         }
63         while(a[flag]==color || a[flag]==0)  //往後找 
64         {
65             if(cnt==n) break;
66             cnt++;
67             flag++;
68             if(flag>n) flag=1;
69         }
70         maxn=max(maxn,cnt);
71     }
72     printf("%d\n",maxn);
73     //system("pause");
74     return 0;
75 }

Section1.1 -- Broken Necklace