hdu 2577 How to Type（DP）

How to Type

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4658    Accepted Submission(s): 2110

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

题意就是每组给一个字符串,问你打出这个字符串须要敲多少次键盘要求打完后CapsLock保持关闭.(注意按一次Shift仅仅能输入一个字母,同一时候在CapsLock开启时使用Shift能输入小写字母). 作为菜鸟。下边是看大神的分析：：：  这题能够直接将连续的(大写和小写同样的)字母依据长度分为长度为1和长度为1以上的来处理,一遍遍历后得出结果(复杂度O(n)),当然也能够使用DP来做. 这里给出DP的做法. 对于每个字符,针对输入前后CapsLock开启关闭状态分类 若输入后CapsLock关闭 假如输入前关闭,则要么直接输入(小写字母),要么用Shift+当前字母输入(大写字母) 假如输入前开启,要么先输入当前字母再按CapsLock(当前字母是大写的),要么先按CapsLock再输入字母(当前字母小写),都是按两次 若输入后CapsLock开启 假如输入前开启,则要么直接输入(大写字母),要么用Shift+当前字母输入(小写字母) 假如输入前关闭,要么先输入当前字母再按CapsLock(当前字母是小写的),要么先按CapsLock再输入字母(当前字母大写),都是按两次 由此能够得出状态转移方程 我们用DP[i][0]来代表输入第i个字母后且CapsLock关闭的情况下须要的最少次数,用DP[i][1]来代表输入第i个字母后且CapsLock开启的情况下须要的最少次数,并定义函数int check(char ch),若ch为大写字母,则返回1,否则返回0,则有 dp[i][0]=min(dp[i-1][0]+check(s[i])+1,dp[i-1][1]+2) dp[i][1]=min(dp[i-1][0]+2,dp[i-1][1]+2-check(s[i])) 特殊的,当i==0时 dp[0][0]=check(s[i])+1 dp[0][1]=2 多练，，，。2015,7,21

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char s[110];
int dp[110][2];
int check(char a)
{
	if(a>='A'&&a<='Z')
		return 1;
	return 0;
}
int main()
{
	int t,i,len;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%s",s);
		len=strlen(s);
		for(i=0;i<len;i++)
		{
			if(i==0)
			{
				dp[i][0]=check(s[0])+1;
				dp[i][1]=2;
			}
			else
			{
				dp[i][0]=min(dp[i-1][0]+check(s[i])+1,dp[i-1][1]+2);
				dp[i][1]=min(dp[i-1][1]+2-check(s[i]),dp[i-1][0]+2);
			}
		}
		printf("%d
",dp[len-1][0]);
	}
	return 0;
}