USACO 1.1 Broken Necklace
题目来源： USACO 1.1
原题目：
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
****** *****
译题：
你有一条由N个红色的，白色的，或蓝色的珠子组成的项链(3<=N<=350)，珠子是随意安排的。 这里是
n=29 的二个例子:
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
图片 A 图片 B
r 代表 红色的珠子
b 代表 蓝色的珠子
w 代表 白色的珠子
第一和第二个珠子在图片中已经被作记号。
图片 A 中的项链可以用下面的字符串表示：
brbrrrbbbrrrrrbrrbbrbbbbrrrrb .
假如你要在一些点打破项链,展开成一条直线，然后从一端开始收集同颜色的珠子直到你遇到一个不
同的颜色珠子，在另一端做同样的事(颜色可能与在这之前收集的不同)。 确定应该在哪里打破项链来收
集到最大多数的数目的子。
例如，在图片 A 中的项链，可以收集到8个珠子,在珠子 9 和珠子 10 或珠子 24 和珠子 25 之间打断
项链。 在一些项链中，包括白色的珠子如图片 B 所示。 当收集珠子的时候，一个被遇到的白色珠子可
以被当做红色也可以被当做蓝色。表现项链的字符串将会包括三符号 r ， b 和 w 。 写一个程序来确
定从一条被给出的项链最大可以被收集珠子数目。
程序名称
beads
输入格式
第 1 行: N, 珠子的数目
第 2 行: 一串度为N的字符串, 每个字符是 r ， b 或 w。
样例输入
(文件 beads.in)
29
wwwbbrwrbrbrrbrbrwrwwrbwrwrrb
输出格式
单独的一行包含从被供应的项链可以被收集的珠子数目的最大值。
样例输出
(文件 beads.out)
View Code
1 /*
2 ID:flyings3
3 PROG:beads
4 LANG:C++
5 */
6 #include <fstream>
7 using namespace std;
8 short n,g=0;
9 short s[350];
10
11 int main()
12 {
13 ifstream fin("beads.in");
14 fin>>n;
15 short i;
16 char t;//取一个字符后停止
17 for (i=0;i<n;i++)
18 {
19 fin>>t;
20 if (t=='r')
21 s[i]=-1;
22 else if (t=='w')
23 s[i]=0;
24 else
25 s[i]=1;
26 }//把字母数字化,红色-1,白色0,蓝色1
27 fin.close();
28 short maxt,j,l,r;
29 for (i=0;i<n;i++)
30 {
31 maxt=0;
32 if (i==0)
33 l=s[n-1];
34 else
35 l=s[i-1];
36 r=s[i];
37 //当i=0（为断点位置，i=0表示点在第一个球之前）的特殊情况，l负责记录左边的颜色,r负责记录右边的颜色
38 if (i>0)
39 j=i-1;
40 else
41 j=n-1;
42 //j向左搜索的起点位置
43 do
44 {
45 if (l==0)
46 l=s[j];
47 maxt++;
48 j--; //往左移一个
49 if (j==-1) //如果到最左了就移到最右
50 j=n-1;
51 }while (((s[j]!=-l)||(s[j]==0))&&(j!=i));
52 //while语句&&前的表示该球为白色或者该球不与前面的球颜色相反（考虑到白色的情况）就继续，后面表示不回到原点就继续
53 if (j!=i) //如果从左没有能数完全部的球
54 {
55 j=i;
56 do
57 {
58 if (r==0)
59 r=s[j];
60 maxt++;
61 j++;
62 if (j==n)
63 j=0;
64 }while((s[j]!=-r)||(s[j]==0)); //不用考虑j==i的问题
65 }
66 else
67 {
68 g=n;
69 break;
70 }
71 if (maxt>g)
72 g=maxt;
73 }
74
75 if (g>n)
76 g=n;
77
78 ofstream fout("beads.out");
79 fout<<g<<endl;
80 fout.close();
81 return 0;
82 }
