0
点赞
收藏
分享

微信扫一扫

hdu2577 How to Type---DP


原题链接: ​​ http://acm.hdu.edu.cn/showproblem.php?pid=2577​​


一:原题内容


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


Hint


 


二:分析理解

 题目大意: 给你一些字符串a~z||A~Z,然后我们按键最少能把他们都打出来,如果开始Cap大写灯是亮的,我们需要打a的话,可以按下cap键,然后再按a,需要两步,但是这下状态变为了大写灯亮,当然我们也可以按下shift,再按下a,需要两步,但是这下状态就变成了大写灯灭。题目要求的是,把这些字母打出来需要的最小按键数目。有一点需要注意,按键完成之后需要使cap键关闭。

解题思路: 用dp[i][1]表示完成i打印Cap灯亮最小的步数,dp[i][0]表示完成i打印Cap灯灭的最小的步数,那么状态转移的时候dp[i][?]只能从dp[i-1][?]中转移,而且选小的,这样就好办了。

三:AC代码

#include<iostream>  
#include<string.h>
#include<algorithm>

using namespace std;

char s[105] = "#";
int dp[105][2];

int main()
{
int test;
scanf("%d", &test);

while (test--)
{
scanf("%s", s + 1);
int len = strlen(s + 1);

dp[0][0] = 0;
dp[0][1] = 1;

for (int i = 1; i <= len; i++)
{
if (s[i] >= 'A'&&s[i] <= 'Z')
{
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);
}
else
{
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);
}
}

printf("%d\n", min(dp[len][0], dp[len][1] + 1));
}


return 0;
}






举报

相关推荐

0 条评论