HDOJ2577 How to Type[DP（两个DP数组）]

How to Type

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1855    Accepted Submission(s): 876

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

 

Author

Dellenge

 

Source

HDU 2009-5 Programming Contest

 

Recommend

lcy

 

 

 

 

 

 

 

 

code:

思路：

开另个数组on[]， off[]。分别默示开Caps Lock灯，和关Caps Lock灯时的最小按键次数

当字符串s[i]为大写时：

　　on[i] = min{ on[i-1] +1， off[i-1] +2} //开灯时直接输入字母，关灯时按shift+字母；

　　off[i] = min{ on[i-1] +2， off[i-1] +2} //开灯时先输字母再关灯，关灯时按shift+字母；

当字符串s[i]为小写时：

　　on[i] = min{ on[i-1] +2， off[i-1] +2} //开灯时按shift+字母， 关灯时按字母再开灯；

　　off[i] = min{ on[i-1] +2， off[i-1] +1} //开灯时关灯再按键，关灯时直接按键；

code:

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
    int on[110],off[110];
    int t;
    char str[110];
    scanf("%d",&t);
    while(t--)
    {
        int i;
        scanf("%s",str);
        on[0]=1;
        off[0]=0;
        for(i=1;i<=strlen(str);i++)
        {
            if(str[i-1]<='Z')
            {
                on[i]=min(on[i-1]+1,off[i-1]+2);
                off[i]=min(on[i-1]+2,off[i-1]+2);
            }
            else
            {
                on[i]=min(on[i-1]+2,off[i-1]+2);
                off[i]=min(on[i-1]+2,off[i-1]+1);
            }
        }
        printf("%d\n",off[i-1]);
    }
    return 0;
}