hdu 2577 DP

原创于 2019-07-31 14:49:11 发布 · 137 阅读

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How to Type

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8642 Accepted Submission(s): 3797

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

这道题有点魔法。开始以为只是一个模拟，后来发现可以按住shift键。理解好后就是一个简单的dp

dp【i】【0】表示当前输入的字母后，状态为小写

dp【i】【1】表示当前输入的字母后，状态为大写

假设当前要输入的a【i】为大写

dp【i】【0】的状态由 dp【i - 1】【0】+ 2（前一个小写状态按住shift 输入）和dp【i】【1】+ 2（前一个状态的大写直接输入字母，再按一下caps）中取小

dp【i】【1】 = min（dp【i - 1】【1】 + 1（直接输入字符），dp【i - 1】【0】 + 2（先按caps 再输入））。

下面是ac代码：

#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <stdio.h>
#include <ctype.h>
#include <bitset>
#define  LL long long
#define  ULL unsigned long long
#define mod 10007
#define INF 0x7ffffff
#define mem(a,b) memset(a,b,sizeof(a))
#define MODD(a,b) (((a%b)+b)%b)
using namespace std;
const int maxn = 2000;
char a[maxn];
int dp[maxn][2];
int main()
{
    int n;
    scanf("%d",&n);
    while(n--){
      scanf("%s",a + 1);
      dp[0][0] = 0;
      dp[0][1] = 1;
      int len = strlen(a + 1);
      for(int i = 1; i <= len; i++){
        if(isupper(a[i])){
          dp[i][0] = min(dp[i - 1][0] + 2,dp[i - 1][1] + 2);
          dp[i][1] = min(dp[i - 1][1] + 1,dp[i - 1][0] + 2);
        }
        else if(islower(a[i])){
          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[len][0],dp[len][1] + 1));
    }

    return 0;
}