USACO1.1.4 Broken Necklace(破碎的项链)

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 beadsLine 2: a string of N characters, each of which is r, b, or w

SAMPLE INPUT (file beads.in)

29wwwbbrwrbrbrrbrbrwrwwrbwrwrrb

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                       ****** *****                       rrrrrb bbbbb  <-- assignments                       5 x r  6 x b    <-- 11 total

题目大意：计算一串由三种不同字符组成的环中，从任意位置起，向左右两旁可拓展的最长距离和。

解题思路：这道题用标准的搜索是O(n^2)的，可以用类似动态规划的方法优化到O(n)，其实用模拟或者枚举来做更为简单。项链是环形的，但我们只要把两个同样的项链放在一块，就把它转换成线性的了。

以下是我的代码：

/*USER:xingwen wangTASK:beadsLANG:C*/#include<stdio.h>int main(){    freopen("beads.in","r",stdin);    freopen("beads.out","w",stdout);    int i,j,n,Max,ans,after[355]={0},before[355]={0};    char a[355];    scanf("%d",&n);    for(i=0;i<=n;i++)    scanf("%c",&a[i]);    a[n+1]='.';a[0]='.';        for(i=1;i<n;i++)    if(a[i]!=a[n]&&(a[n]!='w'&&a[i]!='w'||a[n]=='w'))    break;    after[n]=i;    for(i=n-1;i>=1;i--)    {            if(a[i]==a[i+1]||a[i+1]=='w')            {                    after[i]=after[i+1]+1;                    if(a[i]!='w'&&a[i]==a[i+after[i]])                    after[i]+=after[i+after[i]];            }             else            after[i]=1;    }    for(i=n;i>1;i--)    if(a[i]!=a[1]&&(a[1]!='w'&&a[i]!='w'||a[1]=='w'))    break;    before[1]=n-i+1;    for(i=2;i<=n;i++)    {           if(a[i]==a[i-1]||a[i-1]=='w')           {                  before[i]=before[i-1]+1;                  if(a[i]!='w'&&a[i]==a[i-before[i]])                  before[i]+=before[i-before[i]];           }            else            before[i]=1;    }    before[0]=before[n];    Max=0;    for(i=1;i<=n;i++)    {           ans=after[i]+before[i-1];           if(a[i-1]=='w'&&i-before[i]>=1)           ans+=before[i-before[i-1]-1];           if(a[i]=='w'&&i+after[i]<=n)           ans+=after[i+after[i]];           if(ans>Max)           Max=ans;    }    if(Max>n)    Max=n;    printf("%d\n",Max);    return 0;}

下面是USACO的官方题解：

/*USER:xingwen wangTASK:beadsLANG:C*/#include<stdio.h>int main(){      freopen("beads.in","r",stdin);      freopen("beads.out","w",stdout);      int i,j,n,Max=0,current,state;      char c,s[800];      scanf("%d%s",&n,s);      for(i=n;i<=2*n-1;i++)      s[i]=s[i-n];      for(i=0;i<n;i++)      {              c=s[i];              if(c=='w')              state=0;              else              state=1;              j=i;              current=0;              while(state<=2)              {                       while(j<n+i&&(s[j]==c||s[j]=='w'))                       {                                 current++;                                 j++;                       }                       state++;                       c=s[j];              }              if(current>Max)              Max=current;      }      printf("%d\n",Max);      return 0;}