cug oj 1479 Treasure Chest Lock （区间dp 思维）
1479: Treasure Chest Lock
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 7 Solved: 5
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Description
Vic has a treasure chest. And there is a lock on the treasure chest. The lock contains a sequence of wheels. Each wheel has the 26 letters of the English alphabet
‘a’ through ‘z’, in order. If you move a wheel up, the letter it shows changes to the next letter in the English alphabet (if it was showing the last letter ‘z’, then it
changes to ‘a’). If you move the wheel down, it changes to show the previous letter in the English alphabet (if it was showing ‘a’, then it changes to ‘z’).
It is also possible to move any subsequence of contiguous wheels in the same direction with only one movement. This has the same effect of moving each of the
wheels within the subsequence on that direction, but saves the effort of doing that one wheel at a time.The lock openswhen the wheels show a secret sequence of l
etters. Currently all wheels are showing the letter ‘a’. Vic wants to know the minimum number of movements you need to open the lock.
Input
The input has several test cases. Each of them is given in exactly one line containing a nonempty string of at most 1000 lowercase letters. The string represents
the secret sequence of letters that opens the lock.
Output
For each test case, output a line containing a single integer with the minimum number of movements to open the lock.
Sample Input
abcxyz
abcdefghijklmnopqrstuvwxyz
aaaaaaaaa
zzzzzzzzz
zzzzbzzzz
*
Sample Output
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define maxn 1005
#define MAXN 200005
#define INF 0x3f3f3f3f
#define mod 20140518
#define eps 1e-6
const double pi=acos(-1.0);
typedef long long ll;
using namespace std;
int n,m;
int up[maxn][40],dn[maxn][40];
char s[maxn];
void solve()
{
int i,j,k;
n=strlen(s+1);
memset(dn,0x3f,sizeof(dn));
memset(up,0x3f,sizeof(up));
dn[1][0]=s[1]-'a';
up[1][0]='a'+26-s[1];
for(i=1; i<n; i++)
{
for(j=0; j<30; j++)
{
if(s[i+1]==s[i])
{
dn[i+1][j]=dn[i][j];
up[i+1][j]=up[i][j];
}
else if(s[i+1]<s[i])
{
if(dn[i][j]<INF)
{
dn[i+1][j]=min(dn[i+1][j],dn[i][j]);
if(j>0) dn[i+1][j-1]=min(dn[i+1][j-1],dn[i][j]);
dn[i+1][j+1]=min(dn[i+1][j+1],dn[i][j]+s[i+1]-s[i]+26);
up[i+1][0]=min(up[i+1][0],dn[i][j]+'a'+26-s[i+1]);
}
if(up[i][j]<INF)
{
up[i+1][j]=min(up[i+1][j],up[i][j]+s[i]-s[i+1]);
if(j>0) up[i+1][j-1]=min(up[i+1][j-1],up[i][j]);
up[i+1][j+1]=min(up[i+1][j+1],up[i][j]+s[i]-s[i+1]+26);
dn[i+1][0]=min(dn[i+1][0],up[i][j]+s[i+1]-'a');
}
}
else
{
if(dn[i][j]<INF)
{
dn[i+1][j]=min(dn[i+1][j],dn[i][j]+s[i+1]-s[i]);
if(j>0) dn[i+1][j-1]=min(dn[i+1][j-1],dn[i][j]);
dn[i+1][j+1]=min(dn[i+1][j+1],dn[i][j]+s[i+1]-s[i]+26);
up[i+1][0]=min(up[i+1][0],dn[i][j]+'a'+26-s[i+1]);
}
if(up[i][j]<INF)
{
up[i+1][j]=min(up[i+1][j],up[i][j]);
if(j>0) up[i+1][j-1]=min(up[i+1][j-1],up[i][j]);
up[i+1][j+1]=min(up[i+1][j+1],up[i][j]+s[i]-s[i+1]+26);
dn[i+1][0]=min(dn[i+1][0],up[i][j]+s[i+1]-'a');
}
}
}
}
int ans=min(dn[n][0],up[n][0]);
printf("%d\n",ans);
}
int main()
{
while(~scanf("%s",s+1))
{
if(s[1]=='*') break ;
solve();
}
return 0;
}
/*
mtezqh
ans 30 e -26-e
*/