HDU 2577 How to Type（dp）

最新推荐文章于 2019-03-12 20:23:32 发布

BBHHTT

最新推荐文章于 2019-03-12 20:23:32 发布

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分类专栏：动态规划

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本文介绍了一种算法，用于计算在特定键盘布局下输入指定字符串所需的最小按键次数，包括大小写转换。通过动态规划的方法，该算法考虑了Caps Lock状态的影响。

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How to Type

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7829 Accepted Submission(s): 3519

Problem Description

Pirates have finished developing the typing software. He called Cathy to test his typing software. She is good at thinking. After testing for several days, she finds that if she types a string by some ways, she will type the key at least. But she has a bad habit that if the caps lock is on, she must turn off it, after she finishes typing. Now she wants to know the smallest times of typing the key to finish typing a string.

Input

The first line is an integer t (t<=100), which is the number of test case in the input file. For each test case, there is only one string which consists of lowercase letter and upper case letter. The length of the string is at most 100.

Output

For each test case, you must output the smallest times of typing the key to finish typing this string.

Sample Input

3

Pirates

HDUacm

HDUACM

Sample Output

8

8

8

Hint

The string “Pirates”, can type this way, Shift, p, i, r, a, t, e, s, the answer is 8.
The string “HDUacm”, can type this way, Caps lock, h, d, u, Caps lock, a, c, m, the answer is 8
The string "HDUACM", can type this way Caps lock h, d, u, a, c, m, Caps lock, the answer is 8

设a[ i ]，b[ i ]两数组分别代表Caps lock关与开时打印第 i 个字母的最少操作数。同时又要分大小写两种情况考虑：

c[ i ]是小写时：

a[i]=min(a[i-1]+1,b[i-1]+2);Caps lock关，小写直接输入+1；Caps lock开，shift +输入，共2次。

b[i]=min(a[i-1]+2,b[i-1]+2);小写直接输入+1，同时打开Caps lock，两次； Caps lock开，shift +输入，共2次。

c[ i ]是大写时：

a[i]=min(a[i-1]+2,b[i-1]+2);Caps lock关，shift +输入，共2次；Caps lock开，输入加关闭Caps lock，共2次。

b[i]=min(a[i-1]+2,b[i-1]+1);打开Caps lock，再输入，两次；直接输入1次

最后要注意b数组b[ 0 ]=1 , 最开始要打开Caps lock，操作一次；b[ len ]+=1 , 最后要关闭Caps lock

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int a[105];
int b[105];
char c[105];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		b[0]=1;
		scanf("%s",c+1);
		int len=strlen(c+1);
		for(int i=1;i<=len;i++)
		{
			if(c[i]>='a'&&c[i]<='z')
			{
				a[i]=min(a[i-1]+1,b[i-1]+2);
				b[i]=min(a[i-1]+2,b[i-1]+2);
			}
			else if(c[i]>='A'&&c[i]<='Z')
			{
				a[i]=min(a[i-1]+2,b[i-1]+2);
				b[i]=min(a[i-1]+2,b[i-1]+1);
			}
		}
		b[len]++;
		if(a[len]<b[len])
		{
			printf("%d\n",a[len]);
		}
		else printf("%d\n",b[len]);
	}
	return 0;
 }