HDU 2577 How to Type（dp+状态标记）

How to Type

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6816    Accepted Submission(s): 3077

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

3PiratesHDUacmHDUACM

 

Sample Output

888
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 8The string "HDUACM", can type this way Caps lock h, d, u, a, c, m, Caps lock, the answer is 8
 

用0和1分别表示大写键的状态 0关 1开

那么用dp就可以模拟每一种状态，取其中最小值

四种情况：自己考虑，很好想

#include <iostream>#include<stdio.h>#include<string.h>#include<cctype>#include<algorithm>#define mem(a,b) memset(a,b,sizeof(a))using namespace std;typedef long long ll;char s[105];int dp[105][2];int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%s",s+1);        mem(dp,0);        int n=strlen(s+1);        if(isupper(s[1]))        {            dp[0][0]=0;            dp[0][1]=1;        }        for(int i=1; i<=n; i++)        {            if(isupper(s[i]))            {                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]+2);            }            else            {                dp[i][0]=min(dp[i-1][0]+1,dp[i-1][1]+2);                dp[i][1]=min(dp[i-1][1]+2,dp[i-1][0]+2);            }        }        printf("%d\n",min(dp[n][0],dp[n][1]+1));    }}