hdu 2577 How to Type

How to Type

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2685    Accepted Submission(s): 1244

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
 

 

题意：用键盘输入一串字符串，字符串有大写字母或小写字母，要求输入完这个字符串时最小的按键盘次数，如果输入完时caps lock键是开着的，必须把它关闭。

思路：设on[i]为打印了i个字符后且caps lock键是on状态的按键次数，off[i]为打印了i个字符后且caps lock键是off状态的按键次数。

          有转移方程：

          当前字符为小写字母时：

          on[i]=min(on[i-1]+2,off[i-1]+2);
          off[i]=min(on[i-1]+2,off[i-1]+1);

          当前字符为大写字母时：

          on[i]=min(on[i-1]+1,off[i-1]+2);
          off[i]=min(on[i-1]+2,off[i-1]+2);

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <stack>
#include <map>


using namespace std;


int main()
{
    int on[105],off[105];
    int t;
    char s[105];
    scanf("%d\n",&t);
    while(t--)
    {
        scanf("%s",s+1);
        on[0]=1;
        off[0]=0;
        int len=strlen(s+1);
        for(int i=1;i<=len;i++)
        {
            if(s[i]>='a'&&s[i]<='z')
            {
                on[i]=min(on[i-1]+2,off[i-1]+2);
                off[i]=min(on[i-1]+2,off[i-1]+1);
            }
            else
            {
                on[i]=min(on[i-1]+1,off[i-1]+2);
                off[i]=min(on[i-1]+2,off[i-1]+2);
            }
        }
        int ans=min(on[len]+1,off[len]);
        printf("%d\n",ans);
    }
    return 0;
}