题目链接

Description

Farmer John has arranged his N (1 ≤ N ≤ 5,000) cows in a row and many of them are facing forward, like good cows. Some of them are facing backward, though, and he needs them all to face forward to make his life perfect.

Fortunately, FJ recently bought an automatic cow turning machine. Since he purchased the discount model, it must be irrevocably preset to turn K (1 ≤ K ≤ N) cows at once, and it can only turn cows that are all standing next to each other in line. Each time the machine is used, it reverses the facing direction of a contiguous group of K cows in the line (one cannot use it on fewer than K cows, e.g., at the either end of the line of cows). Each cow remains in the same location as before, but ends up facing the opposite direction. A cow that starts out facing forward will be turned backward by the machine and vice-versa.

Because FJ must pick a single, never-changing value of K, please help him determine the minimum value of K that minimizes the number of operations required by the machine to make all the cows face forward. Also determine M, the minimum number of machine operations required to get all the cows facing forward using that value of K.

Input

Line 1: A single integer: N

Lines 2..N+1: Line i+1 contains a single character, F or B, indicating whether cow i is facing forward or backward.

Output

Line 1: Two space-separated integers: K and M

Sample Input

7

B

B

F

B

F

B

B

Sample Output

3 3

Hint

For K = 3, the machine must be operated three times: turn cows (1,2,3), (3,4,5), and finally (5,6,7)

分析:

从第一个开始找如果当前为B 就需要改变 改变i到i+k-1的位置

一直找到n-k+1处 之后判断从n-k+2到n是否满足了题意 注意不要真的去修改

用一个sum值标记前面k-1个修改了多少次用来决定当前的是否要修改 利用尺取法 即前后推进

代码:

#include<stdio.h>
#include<string.h>
int a[5100],n,flag[5100];
int solve(int k)
{
int i;
memset(flag,0,sizeof(flag));//flag[i]表示区间[i,i+k-1] 是否需要翻转
int sum=0,cnt=0;//前k-1个转变的次数
for(i=1; i<=n-k+1; i++) //sum记录走到当前i,其前面k-1个翻转了多少次
{
if(i-k>=1)///因为每次翻转的是k个,
{
sum-=flag[i-k];
}
if(a[i]==0&&sum%2==0)///如果是B 且前面翻转了偶数次 仍旧需要翻转
{
flag[i]=1;
sum+=flag[i];
cnt++;
}
else if(a[i]==1&&sum%2==1)///如果是F 且前面翻转了奇数次
{
flag[i]=1;
sum+=flag[i];
cnt++;
}
// printf("i=%d flag[i]=%d\n",i,flag[i]);
} for(i; i<=n; i++)
{
if(i-k>=1)
{
sum-=flag[i-k];
}
///这就相当于还没有翻转好
if(sum%2==0&&a[i]==0) return -1;
else if(sum%2==1&&a[i]==1) return -1;
}
return cnt;
} int main()
{
int i,k,mn;
char s[2];
while(scanf("%d",&n)!=EOF)
{
mn=100010000;
for(i=1; i<=n; i++)
{
scanf("%s",s);
if(s[0]=='B') a[i]=0;
else if(s[0]=='F') a[i]=1;
}
// for(int i=1;i<=n;i++)
// printf("%d",a[i]);
// printf("\n");
k=1;
for(i=1; i<=n; i++)///i是指对应的间距
{
int mid=solve(i);///mid是指在相应的间距下的操作次数
///printf("k=%d,cnt=%d\n",i,mid);
if(mid==-1) continue;///在这个间距k下不能满足情况
if(mn>mid)
{
mn=mid;
k=i;
}
}
printf("%d %d\n",k,mn);
}
return 0;
}

POJ 3276 Face The Right Way (尺取法)的更多相关文章

  1. POJ 3061 (二分+前缀和or尺取法)

    题目链接: http://poj.org/problem?id=3061 题目大意:找到最短的序列长度,使得序列元素和大于S. 解题思路: 两种思路. 一种是二分+前缀和.复杂度O(nlogn).有点 ...

  2. POJ 3320 Jessica's Reading Problem 尺取法

    Description Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The fina ...

  3. poj 3320 jessica's Reading PJroblem 尺取法 -map和set的使用

    jessica's Reading PJroblem Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9134   Accep ...

  4. poj 2566Bound Found(前缀和,尺取法)

    http://poj.org/problem?id=2566: Bound Found Time Limit: 5000MS   Memory Limit: 65536K Total Submissi ...

  5. POJ 3320 Jessica's Reading Problem 尺取法/map

    Jessica's Reading Problem Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7467   Accept ...

  6. POJ 3320 尺取法,Hash,map标记

    1.POJ 3320 2.链接:http://poj.org/problem?id=3320 3.总结:尺取法,Hash,map标记 看书复习,p页书,一页有一个知识点,连续看求最少多少页看完所有知识 ...

  7. POJ 3320 Jessica‘s Reading Problem(哈希、尺取法)

    http://poj.org/problem?id=3320 题意:给出一串数字,要求包含所有数字的最短长度. 思路: 哈希一直不是很会用,这道题也是参考了别人的代码,想了很久. #include&l ...

  8. 尺取法 poj 2566

    尺取法:顾名思义就是像尺子一样一段一段去取,保存每次的选取区间的左右端点.然后一直推进 解决问题的思路: 先移动右端点 ,右端点推进的时候一般是加 然后推进左端点,左端点一般是减 poj 2566 题 ...

  9. POJ 3320 尺取法(基础题)

    Jessica's Reading Problem Description Jessica's a very lovely girl wooed by lots of boys. Recently s ...

随机推荐

  1. C#之WCF入门1—简单的wcf例子

    第一步:创建一个空的解决方案,新建一个WCF服务应用程序项目(使用默认名字) 来模拟服务端,新建一个控制台应用程序项目(名称改为 ConsoleApp)来模拟客户端. 第二步:简单分析WcfServi ...

  2. 【week2】 累计进度条、psp、饼图

    每周例行报告 本周PSP 类别 任务 开始时间 结束时间 被打断时间 总计工作时间 2016年9月9日 读书 构建之法-5.6章 19:00 20:00 0 60min 2016年9月10日 看博客 ...

  3. 【重读MSDN之ADO.NET】ADO.NET连接

    连接到ADO.NET中的数据源 在 ADO.NET 中,通过在连接字符串中提供必要的身份验证信息,使用 Connection 对象连接到特定的数据源.使用的 Connection 对象取决于数据源的类 ...

  4. [剑指Offer] 57.二叉树的下一个结点

    题目描述 给定一个二叉树和其中的一个结点,请找出中序遍历顺序的下一个结点并且返回.注意,树中的结点不仅包含左右子结点,同时包含指向父结点的指针. /* struct TreeLinkNode { in ...

  5. AC自动机裸题

    HDU 2222 Keywords Search 模板题.对模式串建立AC自动机然后在trie树上找一遍目标串即可. # include <cstdio> # include <cs ...

  6. 【bzoj3672】[Noi2014]购票 斜率优化dp+CDQ分治+树的点分治

    题目描述  给出一棵以1为根的带边权有根树,对于每个根节点以外的点$v$,如果它与其某个祖先$a$的距离$d$不超过$l_v$,则可以花费$p_vd+q_v$的代价从$v$到$a$.问从每个点到1花费 ...

  7. C# 利用FTP自动下载xml文件后利用 FileSystemWatcher 监控目录下文件变化并自动更新数据库

    using FtpLib; using System; using System.Collections.Generic; using System.ComponentModel; using Sys ...

  8. BZOJ3124 SDOI2013直径

    本以为必有高论,结果是个思博题.随便找一条直径,最后答案肯定是这条直径上的连续一段,如果某分支长度等于直径上某端的长度这一端都要被剪掉. #include<iostream> #inclu ...

  9. Javascript 中 == 和 === 区别是什么?

    Javascript 中 == 和 === 区别是什么? 作者:Belleve链接:https://www.zhihu.com/question/31442029/answer/77772323来源: ...

  10. BZOJ2434:[NOI2011]阿狸的打字机——题解

    https://www.lydsy.com/JudgeOnline/problem.php?id=2434 https://www.luogu.org/problemnew/show/P2414 打字 ...