S 1.1 beads C程序

来源：互联网发布：发货明细清单软件编辑：程序博客网时间：2024/05/11 23:19

题目连接：

http://ace.delos.com/usacoprob2?a=xWwmrH0ntr1&S=beads

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 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)

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

题目翻译：

题目描述

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数目的珠子。
　　例如，在图片 A 中的项链中，在珠子 9 和珠子 10 或珠子 24 和珠子 25 之间打断项链可以收集到8个珠子。

白色珠子什么意思?

   在一些项链中还包括白色的珠子(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。

　　表现含有白珠项链的字符串将会包括三个符号 r ， b 和 w 。
　　写一个程序来确定从一条被给出的项链可以收集到的珠子最大数目。

[编辑]程序名称

beads

[编辑]输入格式

第 1 行: N, 珠子的数目
第 2 行: 一串长度为N的字符串, 每个字符是 r ， b 或 w。

[编辑]样例输入

(文件 beads.in)

29 wwwbbrwrbrbrrbrbrwrwwrbwrwrrb

[编辑]输出格式

单独的一行最大可能取得的珠子数

[编辑]样例输出

(文件 beads.out)

题目求解：

/*
ID:smilecl1
LANG:C
TASK:beads
*/

/*
* TIME：2012/11/1
* MODIFY: 2012/11/8
* EMAL: SmileCloud201@gmail.com
* NAME: LiYun
*/

/*
* 思路：
* 将项链从珠子1与最后一个珠子之间截断看成一条线，为了避免环形问题导致的错误，将两条同样的项链连成一条来计算.
* 注意：当做直线来看，n个珠子，就有n+1 个截断点。我在这个问题上犯过错误！！
*/

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define max_lenth 400

int max(int i, int j)
{
    int m, n;
    m = i;
    n = j;
    return m >= n ? m : n;
}

int min(int i, int j)
{
    int m,n;
    m = i;
    n = j;
    return m <=n ? m : n;
}

int main()
{
FILE *fin;
FILE *fout;

    int i = 0;
    int result = 0;
    int n = 0;
    char temp[max_lenth] = {0};
    char necklace[2 * max_lenth] = {0};
    int red_left[2 * max_lenth] = {0};
    int red_right[2 * max_lenth] = {0};
    int blue_left[2 * max_lenth] = {0};
    int blue_right[2 * max_lenth] = {0};

    fin=fopen("beads.in","r");
    fout=fopen("beads.out","w");
    fscanf(fin, "%d%s", &n, temp);
    strcpy(necklace, temp);
    strcat(necklace, temp);

    /* 从necklace[0]的前面开始切断并计左边珠子的数目 */
    for (i = 0; i != 2 * n; ++i) /*注意red_left blue_left都有2*n+1 个值 */
    {
        if (necklace[i] == 'r')
        {
            red_left[i + 1]=red_left[i] + 1;
            blue_left[i + 1]=0;
        }
        if (necklace[i] == 'b')
        {
            red_left[i + 1] = 0;
            blue_left[i + 1] = blue_left[i] + 1;
        }
        if (necklace[i] == 'w')
        {
            red_left[i + 1] = red_left[i] + 1;
            blue_left[i + 1] = blue_left[i] + 1;
        }
    }

    /* 从最后一个珠子开始计算右边珠子的数目 */
    for (i= 2 * n-1; i > 0; --i) /*注意red_right blue_right都有 2*n+1 个值 */
    {
        if (necklace[i] == 'r')
        {
            red_right[i] = red_right[i + 1] + 1;
            blue_right[i] = 0;
        }
        if (necklace[i] == 'b')
        {
            red_right[i] = 0;
            blue_right[i] = blue_right[i + 1] + 1;
        }
        if (necklace[i] == 'w')
        {
            red_right[i] = red_right[i + 1] + 1;
            blue_right[i] = blue_right[i + 1] + 1;
        }
    }
    for(i = 0; i != 2 * n; ++i)
    {
        result = max( result, max( blue_left[i], red_left[i] ) + max( blue_right[i], red_right[i]) );
    }
    result = min(result, n);
    fprintf( fout, "%d\n", result );
    return 0;
}