HDU 2577 分情况多维DP

How to Type

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6787    Accepted Submission(s):
3057

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

类似于上次写的那道可以上下右走方格的题目，也是常见的一类DP，有一种分层的思想!

对于此题就在于Caps键在DP的过程中可能处于开/关两种，我们不妨多开一维0表示Caps键处于关闭1表示处于开启状态.

直接地推1Ahhh,还是在喝了酒之后好开森>_<

#include<bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
int dp[105][2];
char s[105];
int solve()
{
int i,j,k,sz=strlen(s);
memset(dp,inf,sizeof(dp));
if(islower(s[0])){
dp[0][0]=1;
dp[0][1]=2;
}
else{
dp[0][0]=2;
dp[0][1]=2;
}
for(i=1;i<sz;++i){
if(islower(s[i])){
dp[i][0]=min(dp[i-1][1]+2,dp[i-1][0]+1);
dp[i][1]=min(dp[i-1][1]+2,dp[i-1][0]+2);
}
else{ //da xie
dp[i][0]=min(dp[i-1][1]+2,dp[i-1][0]+2);
dp[i][1]=min(dp[i-1][1]+1,dp[i-1][0]+2);
}
}
return min(dp[sz-1][0],dp[sz-1][1]+1);
}
int main()
{
int t,n,m,i,j;
cin>>t;
while(t--){cin>>s;
cout<<solve()<<endl;
}
return 0;
}