HDOJ2577 DP简单入门题

最新推荐文章于 2020-11-20 18:27:03 发布

最新推荐文章于 2020-11-20 18:27:03 发布 · 406 阅读

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#acm #dp #算法

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本文介绍了一种算法问题，旨在找到完成输入字符串所需的最小按键次数，考虑到大小写敏感及Caps Lock键的影响。通过动态规划的方法，文章展示了如何计算不同类型字符的最佳按键方案。

How to Type

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7194 Accepted Submission(s): 3249

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

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状态容易区分，要么按caps键，要么不按。dp[i][0]表示不按，dp[i][1]表示按。

#include <iostream>
#include <ctime>
#include <cstring>
using namespace std;

const int maxn = 105;
char s[105];
int len;
int dp[maxn][2];
int main(){
    std::ios::sync_with_stdio(false);
    int n;
    cin >> n;
    while (n--) {
        cin >> s;
        len = strlen(s);
        memset(dp,0,sizeof(dp));
        if (s[0]>='A' && s[0] <='Z') dp[0][0]=dp[0][1]=2;
        else {
                dp[0][0]=1;
                dp[0][1]=2;//因为总是按caps键，所以这里加2
        }

        for (int i=1;i<len;i++){
            if ( s[i]>='A' && s[i]<='Z') {
                 dp[i][1]=min(dp[i-1][0]+2,dp[i-1][1]+1);
                 dp[i][0]=min(dp[i-1][0]+2,dp[i-1][1]+3); //dp[i-1][1]+3因为先关掉caps键，然后按shift+字母
            }
            else if (s[i]>='a' && s[i]<='z'){
                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);
            }
        }
        if (s[len-1]>='A' && s[len-1]<='Z') dp[len-1][1]++;//关掉caps键
        cout<<min(dp[len-1][0],dp[len-1][1])<<endl;
    }
    return 0;
}