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)
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 ****** *****
题目大意：有一条完整的项链，有三种颜色珠子构成，red，blue，white，现在要求从某一处断开，分别从断开处求连续颜色珠子的个数，用cnt[]保存，求他们的和的最大值，其中white可以按照需要当做red或blue来用
算法分析：模拟加枚举，先将连续相同的珠子合并成一个节点，这样就构成相邻两个节点异色，枚举每个节点作为最后求和的两个节点的首节点
flag[]用来标记当前节点的颜色
0 -- blue
1 -- red
2 -- white
处理的时候用cur表示当前节点的颜色，遇到white，cur不变，当总的颜色的出现次数大于2时，退出。
注意：第一个节点和最后一个节点也是相邻的，需要另外处理一下
代码
/*
ID: qqxinre1
LANG: C
TASK: beads
*/
#include<stdio.h>
#include<stdlib.h>
#define NN 355
void main()
{
FILE* fin = fopen("beads.in", "r");
FILE* fout = fopen("beads.out", "w");
int N, index, ch, i, ans, cur, sum, j, num;
int flag[NN], cnt[NN];
char str[NN];
fscanf(fin, "%d%s", &N, str);
index = -1;
for (i = 0; i < N; i++){
if (str[i] == 'b')
ch = 0;
else if (str[i] == 'r')
ch = 1;
else
ch = 2;
if (i == 0 || str[i] != str[i - 1]){
cnt[++index] = 1;
flag[index] = ch;
}
else
cnt[index]++;
}
if (index == 0)
{
fprintf(fout, "%d\n", cnt[0]);
exit(0);
}
if (flag[0] == flag[index])
cnt[0] += cnt[index--];
ans = 0;
for (i = 0; i <= index; i++){
cur = -1;
sum = 0;
for (j = i; j <= index; j++){
if (flag[j] == 2){
sum += cnt[j];
continue;
}
if (cur == -1){
cur = flag[j];
num = 1;
}
else{
if (cur != flag[j]){
cur = flag[j];
num ++;
if (num > 2)
break;
}
}
sum += cnt[j];
}
if (j > index){
for (j = 0; j < i; j++){
if (flag[j] == 2){
sum += cnt[j];
continue;
}
if (cur == -1){
cur = flag[j];
num = 1;
}
else{
if (cur != flag[j]){
cur = flag[j];
num ++;
if (num > 2)
break;
}
}
sum += cnt[j];
}
}
if (sum > ans)
ans = sum;
}
fprintf(fout, "%d\n", ans);
exit(0);
}
