hdu 2577 How To Type

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2577

 

题目描述：

 

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

 

 

 

题目大意：

给你一个字符串，初始状态的Caps Lock灯是灭的，在要求最后状态Caps Lock灯也是灭的情况下，问最少要敲多少次键盘才能把这个字符串打出来

 

题目分析：

这里我们用dp[i][0]代表敲完第i个结束灯是灭的，dp[i][1]代表敲完第i个结束灯是开的

状态转移方程：

if(s[i]>='a'&&s[i]<='z')
            {
                dp[i][0]=min(dp[i-1][0]+1,dp[i-1][1]+2);
                dp[i][1]=min(dp[i-1][0]+2,dp[i-1][1]+2);
            }
else
            {
                dp[i][0]=min(dp[i-1][0]+2,dp[i-1][1]+2);
                dp[i][1]=min(dp[i-1][0]+2,dp[i-1][1]+1);
            }

 

AC代码：

 

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iomanip>
#include<algorithm>
using namespace std;

int dp[105][2];
int T;
char s[105];
int l;

int main()
{
    cin>>T;
    while(T--)
    {
        //dp[i][0]代表敲完第i个结束灯是灭的，dp[i][1]代表敲完第i个结束灯是开的
        dp[0][0]=0;//关灯初值
        dp[0][1]=2;//开灯初值，这里是2的原因，可以理解为假设字符串就是0个字符的话，那么刚开始要按一次使他变亮，然后又因为结束要求灯是灭的，所以还需要再按一次使他变黑
        cin>>s+1;
        l=strlen(s+1);
        for(int i=1;i<=l;i++)
        {
            if(s[i]>='a'&&s[i]<='z')
            {
                dp[i][0]=min(dp[i-1][0]+1,dp[i-1][1]+2);
                dp[i][1]=min(dp[i-1][0]+2,dp[i-1][1]+2);
            }
            else
            {
                dp[i][0]=min(dp[i-1][0]+2,dp[i-1][1]+2);
                dp[i][1]=min(dp[i-1][0]+2,dp[i-1][1]+1);
            }
        }
        
        printf("%d\n",min(dp[l][0],dp[l][1]+1));//因为要求结束以后灯是灭的，所以+1
    }
    return 0;
}