hdu2577—How to Type（dp）-优快云博客

本文介绍了一种算法，用于计算输入特定字符串时所需的最小按键次数，考虑到大小写锁定的状态变化。通过动态规划方法，文章详细解释了如何确定最佳按键顺序以减少总的按键数量。

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题目链接：传送门

How to Type

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6811 Accepted Submission(s): 3075

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

解题思路：打第i个字符的Caps状态可以是开or关，用dp[i][0]表示关，dp[i][1]表示开(存储打第i个字符时在某Caps状态下最少需要打多少个字符)，则dp[i][0]和dp[i][1]可以由前面的dp[i-1][0]和dp[i-1][1]决定。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <stack>
#include <queue>

using namespace std;

typedef long long ll;
typedef pair<int,int>PA;
const int N = 109;
const int M = 1000;
const int INF = 0x3fffffff;

int dp[N][0];
//dp[i][0]:将要打第i个字符时Caps关闭状态下最少要打多少个字符
//dp[i][1]:将要打第i个字符时Caps打开状态下最少要打多少个字符

int main()
{
    int T;
    scanf("%d",&T);
    while( T-- ){
        string st;
        cin >> st;
        dp[0][0] = 0;
        dp[0][1] = 1;
        int len = st.length();
        for( int i = 0 ; i < len ; ++i ){
            if( st[i]>='a'&&st[i]<='z' ){
                dp[i+1][0] = min( dp[i][0]+1 , dp[i][1]+2 );
                dp[i+1][1] = min( dp[i][0]+2 , dp[i][1]+2 );
            }
            if( st[i]>='A'&&st[i]<='Z' ){
                dp[i+1][0] = min( dp[i][0]+2 , dp[i][1]+2 );
                dp[i+1][1] = min( dp[i][0]+2 , dp[i][1]+1 );
            }
        }
        //dp[len][1]+1:Caps键打开后最后要关闭
        int ans = min( dp[len][0] , dp[len][1]+1 );
        printf("%d\n",ans);
    }
    return 0;
}