题目链接:HDOJ 2577
题意:打出所给的字符串,大小写都有,并且最终Caps Lock 要关闭,求最少的按键数
题解:简单的dp方程,考虑Caps Lock on 和in的情况dp就可
代码:
#include<cstdio>
#include<queue>
#include<cstring>
#include<string>
#include<stack>
using namespace std;
#define M(a) memset(a,0,sizeof(a))
#define Max(a,b) ((a>b)?a:b)
#define Min(a,b) ((a<b)?a:b)
#define debug 0
const int maxn = 100 + 5;
char str[maxn]; int word[maxn];
int dp[maxn][2];
void Do()
{
int len = strlen(str);
for (int i = 1; i <= len; i++)
if (isupper(str[i - 1])) //判断大小写
word[i] = 1;
int num;
if (word[1]) //初始化第一个字符type
{
dp[1][0] = 2; //dp[i][0]表示在Caps Lock关的状态情况打出第i个字符的最优解
dp[1][1] = 2; //dp[i][1]表示在Caps Lock开的状态情况打出第i个字符的最优解
}
else
{
dp[1][0] = 1;
dp[1][1] = 3;
}
for (int i = 2; i <= len; i++)
{
if (word[i])
{
dp[i][0] = Min(dp[i - 1][0] + 2, dp[i - 1][1] + 3);//注意上一状态Caps Lock要先转Caps Lock再type
dp[i][1] = Min(dp[i - 1][0] + 2, dp[i - 1][1] + 1);
}
else
{
dp[i][0] = Min(dp[i - 1][0] + 1, dp[i - 1][1] + 2);
dp[i][1] = Min(dp[i - 1][0] + 3, dp[i - 1][1] + 2);
}
}
printf("%d\n", Min(dp[len][0], dp[len][1] + 1));//最终Caps Lock要关闭
}
int main()
{
#if debug
freopen("in.txt", "r", stdin);
#endif//debug
int T;
while (~scanf("%d", &T))
{
while (T--)
{
M(dp);
M(word);
scanf("%s", str);
Do();
}
}
return 0;
}