HDU 2577 How to Type (线性dp)

How to Type

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4616    Accepted Submission(s): 2084

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



题目链接：

pid=2577">http://acm.hdu.edu.cn/showproblem.php?pid=2577



题目大意：坑比题，给一些有大写和小写组成的字符串，问用键盘打出它们的最少按键次数

按键规则：对于每个小写字母。加上shift可使其变成大写。注意shift不能够一直按着不放，还能够开大写锁Caps lock键，按完这键输入就变成大写，此时再按shift。则打出小写（mac的键盘貌似不是这样），注意每次大写锁，最后必须把它关上



题目分析：知道题以后就非常easy了dp[i][0]和dp[i][1]分别表示到第i个字符大写锁关，开着时要按的最少次数，则有4种情况

假设到第i个字符为小写且此时开着大写锁，则它前一个必定是大写，由于假设前一个是小写，当前也是小写。我显然不须要开大写锁

dp[i][1] = dp[i - 1][1] + 2

假设到第i个字符为小写且此时没开大写锁，直接打即可了

dp[i][0] = dp[i - 1][0] + 1

假设到第i个字符为大写且此时开着大写锁。则为前一个开着大写锁的情况加1和前一个没开大写锁的情况加2的最小值，表示开锁

dp[i][1] = min(dp[i - 1][1] + 1, dp[i - 1][0] + 2)

假设到第i个字符为大写且此时没开大写锁。则为前一个没开大写锁的情况加2(按shift)和前一个开着大写锁的情况加2的最小值。表示关锁

dp[i][0] = min(dp[i - 1][0] + 2, dp[i - 1][1] + 2)

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const MAX = 105;
char s[MAX];
int dp[MAX][2];

bool judge(char ch)
{
    if(ch >= 'A' && ch <= 'Z')
        return true;
    return false;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%s", s + 1);
        int len = strlen(s + 1);
        memset(dp,0, sizeof(dp));
        dp[0][1] =  1;
        for(int i = 1; i <= len; i++)
        {
            if(judge(s[i]))
            {
                dp[i][1] = min(dp[i - 1][1] + 1, dp[i - 1][0] + 2);
                dp[i][0] = min(dp[i - 1][0] + 2, dp[i - 1][1] + 2);
            }
            else
            {
                dp[i][1] = dp[i - 1][1] + 2;
                dp[i][0] = dp[i - 1][0] + 1;
            }
        }
        printf("%d\n", min(dp[len][0], dp[len][1] + 1));
    }
}