1.1 beads

拾取珠子

Broken Necklace

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. The string that represents this configuration will include the three symbols r, b and w.

Write a program to determine the largest number of beads that can be collected from a supplied necklace.
PROGRAM NAME: beads
INPUT FORMAT
Line 1:     N, the number of beads
Line 2:     a string of N characters, each of which is r, b, or w
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.

wwwbbrwrbrbrrbrbrwrwwrbwrwrrb wwwbbrwrbrbrrbrbrwrwwrbwrwrrb
                       ****** *****
/*ID: xdzhbin1LANG: C++TASK:beads*/#include <fstream>#include <vector>using namespace std;//把三种颜色定义为常量const char red = 'r';const char blue ='b';const char white = 'w';//返回下一个元素的索引，项链是个循环链表，所以最后一个元素的下一个元素就是第一个元素int next(int index, int size){return (index+1)%size;}//返回一个元素的前一个元素，第一个元素的前一个元素就是最后一个元素int forward(int index, int size){return (index-1+size)%size;}//返回从索引index出切断项链，可以搜集到的珠子数目int collect(vector<char> &v, int index){int count = 0;//计数int i=index,j;//i用作向后查看，j用作向前查看char color;//当前可以搜集起来珠子的颜色int size = v.size();//总共的珠子数量int whistle;//后向收集的哨子，记录后向收集的位置，防止重复收集//向后统计，并得到接受的颜色while(v[i]==white){i=next(i,size);count++;if(i == index)return size;//全是白色球的情况}color = v[i];while(v[i]==white || v[i]==color){i=next(i,size);count++;if(i == index)return size;//已经全部收集起来}whistle = forward(i,size);//向前统计，并得到接受的颜色//向前统计，肯定不会出现把全部项链都收集起来的情况//因为如果可以全部收集起来，向后统计就已经全部收集起来了j = forward(index,size);while(v[j]==white){j=forward(j,size);count++;}color = v[j];while(v[j]==white || v[j]==color){j = forward(j,size);count++;if(j == whistle) break;}return count;}int main(){ifstream in("beads.in",ios_base::in);ofstream out("beads.out",ios_base::out);int num_breads;//记录项链的珠子数目vector<char> vbreads;//记录各个珠子in>>num_breads;char ch;while(in>>ch){vbreads.push_back(ch);}int i,max = 0,sum;for(i=0; i<vbreads.size(); i++){sum = collect(vbreads,i);//从i处截断可以收集的最大珠子数目if(sum > max){max= sum;}}out<<max<<endl;in.close();out.close();return 0;}