# USACO 1.1 Broken Necklace 破碎的项链 解题报告

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

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.

INPUT FORMAT
Line 1:  N, the number of beads
Line 2:  a string of N characters, each of which is r, b, or w

29
wwwbbrwrbrbrrbrbrwrwwrbwrwrrb

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

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

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

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

29
wwwbbrwrbrbrrbrbrwrwwrbwrwrrb

11

========================= 华丽的分割线 =========================
这个题目我用的思路是标程里面的一个代码, 用a表示前一段的连续长度, b表示当前段的连续长度, w表示在当前出现的w的连续数目.. 对于对于循环的话, 就把字符串f再复制一份到f后面….
啥~? 没看懂~? 看样子对于程序员来说最好的语言是代码:

/*
ID: zqy11001
LANG: C
*/
#include <stdio.h>
#define getint(i) scanf(%d\\n, &n)

int n;
char f[701];

int main(void)
{
int i, limit;
int a = 0, b = 0, w = 0;
char c = \'0\';
int m = 0;
getint(n);
limit = 2 * n;
fgets(f, 351, stdin);
memcpy(f + n, f, n);
for(i = 0; i < limit; i++){
if(f[i] == \'w\'){
b++;
w++;
}else if(f[i] == c){
b++;
w = 0;
}else{
if(b + a > m){
m = b + a;
}
a = b - w;
b = w + 1;
w = 0;
c = f[i];
}
}
if(a + b > m){
m = b + a;
}
printf(%d\\n, m > n ? n : m);
return 0;
}