2577 HUD (动态规划)_hud acm动态规划-优快云博客

本文链接：https://blog.youkuaiyun.com/u014737310/article/details/38308269

本文通过描述解决海盗使用打字软件时的优化问题，深入探讨了动态规划（DP）算法的应用。通过实例分析，展示了如何利用DP解决实际问题，并提供了一段C++代码实现。重点介绍了状态转移方程的构建，以及如何通过最小化打字次数来优化输入过程。

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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

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

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 105

int 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;
}