HDU2577 How to Type (动态规划)

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本文介绍了一种算法，用于计算输入特定字符串时按键的最少次数，考虑到大小写转换的不同方式，并确保Caps Lock最终处于关闭状态。通过动态规划方法实现了这一目标。

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How to Type

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6440 Accepted Submission(s): 2909

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

题目要求：求输入一个字符串按键的最少次数，大写字母可以按shift + 字符也可以先按cops lock键再按字符，最后输入完毕时cops lock必须处于关闭状态。

思路：数组dpa[]表示关灯的状况，dpb[]表示开灯的状况，也因此dpa[0]初始化为0，dpb[0]初始化为1。当扫描的第i个字符时，可以得出：

如果a[i]是小写字母时， dpa[i + 1] = min(dpa[i] +1, dpa[i] + 2) // 表示想要该状态为关灯状态，前者直接输入字符即可，后者先关灯再输入。

dpb[i + 1] = min(dpa[i] + 2, dpb[i] + 2); //表示想要该状态为开灯状态，前者先输入再开灯，后者按shift + 字符输入。

如果a[i]是大写字母时，

dpa[i + 1] = min(dpa[i] + 2,dpb[i] + 2),表示想要该状态为关灯状态，前者按shift + 字符输入，后者先输入再关灯，，dpb[i] = min(dpa[i] + 2, dpb[i] + 1);表示想要该状态为开灯状态，前者先开灯再输入，后者直接输入即可，,最后输出结果取最小值即可。

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

int dpa[105],dpb[105];
int t;
char a[105];

int main(){
	cin >> t;
	
	while(t --){
		cin >> a;
		int len = strlen(a);
		
		dpa[0] = 0;		dpb[0] = 1;
		
		for(int i = 0; i < len; i ++){
			if(a[i] >= 'a' && a[i] <= 'z'){
				dpa[i + 1] = min(dpa[i ] + 1, dpb[i ] + 2);
				dpb[i + 1] = min(dpa[i ] + 2, dpb[i ] + 2);
			}
			else if(a[i] >= 'A' && a[i] <= 'Z'){
				dpa[i + 1] = min(dpa[i ] + 2, dpb[i ] + 2);
				dpb[i + 1] = min(dpa[i ] + 2, dpb[i ] + 1);
			}
		}
		cout << min(dpa[len], dpb[len] + 1) << endl;
	}
	return 0;
}