2577 HUD (动态规划)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2577

看别人代码做了这个题，感觉dp好强大，很震撼，哎，我还是太年轻了，学的太少了

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
 

 

 定义两个数组，一个是始终开的状态，一个是始终关的状态，然后就相互转化去改状态最小值，千万要记得可以用 Shift   键啊！！！！

解释在代码中，

#include<iostream>#include<cstdio>#include<cstring>using namespace std;#define N 105int a[N];  //Num  Lock 键始终开int b[N];  //Num  Lock 键始终关inline int min(int a,int b){    return a<b?a:b;}int main(){    char c[N];    int t,i;    scanf("%d",&t);    while(t--)    {        scanf("%s",c+1);  //方便而已，自己猜测吧        int len=strlen(c+1);  //如果是strlen(c),len=0;        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        a[0]=1;  //打开Num  Lock 键        for(i=1;i<=len;i++)            if(c[i]>='a'&&c[i]<='z')            {                a[i]=min(a[i-1]+2,b[i-1]+2);                        //Shift+字母    字母+Num  Lock 键                b[i]=min(a[i-1]+2,b[i-1]+1);                     //Num Lock键+字母   字母                 }            else            {                a[i]=min(a[i-1]+1,b[i-1]+2);  //同理                b[i]=min(a[i-1]+2,b[i-1]+2);            }            a[len]++; //关掉Num Lock键            int ans=min(a[len],b[len]);            printf("%d\n",ans);    }    return 0;}