HDU 2577

How to Type

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6072    Accepted Submission(s): 2729

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

 

Recommend

lcy

题意：

问你打一串字母需要最少的按几次键盘。caps lock亮着时按shift可以打印小写字母！

思路：

每一次打字母时的总按键数都可以由前一次的推断出来，

当为大写字母时:  st[i+1][0]=min(st[i][0]+2,st[i][1]+2);
                       st[i+1][1]=min(st[i][0]+2,st[i][1]+1);

小写字母时：      st[i+1][0]=min(st[i][0]+1,st[i][1]+2);
                       st[i+1][1]=min(st[i][0]+2,st[i][1]+2);

st[0][0],st[0][1]分别初始化为0和1.

最后如果capslock亮，还要再加1.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-);
const int MAXN = ;
int t, f[][];
char s[];
int main()
{
    scanf("%d", &t);
    while (t--) {
        scanf("%s", s + );
        int len = strlen(s + );
        f[][] = ; f[][] = ;
        for (int i = ; i <= len; ++i) {
            if (s[i] >= 'a' && s[i] <= 'z') {
                f[i][] = min(f[i - ][] + , f[i - ][] + );
                f[i][] = min(f[i - ][] + , f[i - ][] + );
            }
            else {
                f[i][] = min(f[i - ][] + , f[i - ][] + );
                f[i][] = min(f[i - ][] + , f[i - ][] + );
            }
        }
        printf("%d\n", min(f[len][], f[len][] + ));
    }
    return ;
}