How to Type - HDU 2588 dp

最新推荐文章于 2019-10-10 19:48:47 发布

原创最新推荐文章于 2019-10-10 19:48:47 发布 · 595 阅读

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本文探讨了如何通过shift和CapsLock键打出含有大小写的字符串，并计算最少需要敲击键盘的次数。通过动态规划方法解决该问题，每一步选择是否开启或关闭CapsLock键以达到最小操作数。

How to Type

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3496 Accepted Submission(s): 1623

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shift和Caps Lock打出含有大小写的字符串，问最少需要敲多少次键盘。注意最后要关闭Caps Lock。

思路：dp[i][0]表示打完第i个字符后Caps Lock为关的最少次数，dp[i][1]表示打完第i个字符后Caps Lock为开的最少次数，转移方程看代码吧。

AC代码如下：

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<string>
using namespace std;
int dp[110][2];
char s[110];
int main()
{ int t,i,j,k,len,ans=0;
  scanf("%d",&t);
  while(t--)
  { scanf("%s",s+1);
    len=strlen(s+1);
    dp[0][0]=0;dp[0][1]=1;
    for(i=1;i<=len;i++)
    { if(islower(s[i]))
      { dp[i][0]=min(dp[i-1][0]+1,dp[i-1][1]+2);
        dp[i][1]=min(dp[i-1][1]+2,dp[i-1][0]+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[len][0],dp[len][1]+1));
  }
}