2019CCPC秦皇岛I题 Invoker(DP)

Invoker

Time Limit: 15000/12000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 117    Accepted Submission(s): 35

Problem Description

In dota2, there is a hero named Invoker. He has 3 basic skills in the game, which are Quas, Wex and Exort. Once he launches a basic skill, he will gain the corresponding element, where Quas gives "Q", Wex gives "W" and Exort gives "E".
Invoker can't have more than 3 elements simultaneously. If he launches a basic skill when he already owns 3 elements, he will get the corresponding element and lose the element he gained the earliest.
As can be seen, there are 10 unordered combinations of 3 elements in 3 types, each represents a special skill, which are as follows:

• Cold Snap: unordered element combination "QQQ", denoted by "Y"

• Ghost Walk: unordered element combination "QQW", denoted by "V"

• Ice Wall: unordered element combination "QQE", denoted by "G"

• EMP: unordered element combination "WWW", denoted by "C"

• Tornado: unordered element combination "QWW", denoted by "X"

• Alacrity: unordered element combination "WWE", denoted by "Z"

• Sun Strike: unordered element combination "EEE", denoted by "T"

• Forge Spirit: unordered element combination "QEE", denoted by "F"

• Chaos Meteor: unordered element combination "WEE", denoted by "D"
• Deafening Blast: unordered element combination "QWE", denoted by "B"

When Invoker owns 3 elements, he can launch the invoking skill, denoted by "R", to gain the special skill according to the elements he currently owns. After invoking, the elements won't disappear, and the chronological order of the 3 elements won't change.
Now given a sequence of special skills, you want to invoke them one by one with using the minimum number of basic skills(Q,W,E) and invoking skill(R). Print the minimum number in a single line.
At the beginning, Invoker owns no elements. And you should re-invoke the special skills even if you have already invoked the same skills just now.

 

 

Input

Input a single line containing a string s (1 ≤ |s| ≤ 100 000) that only contains uppercase letters in {B, C, D, F, G, T, V, X, Y, Z}, denoting the sequence of special skills.

 

 

Output

Output a single line containing a positive integer, denoting the minimum number of skills to launch.

 

 

Sample Input

XDTBVV

 

 

Sample Output

15

Hint

One possible scheme is QWWREERERWQRQRR.

 

 

Source

642ccpcQHD

 

---

参考代码：

#include<bits/stdc++.h>
using namespace std;
const int size=1e5+5;
const int inf=0x3f3f3f3f;
char ss[size];
int dp[size][5][5];
int ans[size];
char s[15][5]={"QQQ","QQW","QQE","WWW","QWW","WWE","EEE","QEE","WEE","QWE"};
int stru[6][3]={{0,1,2},{0,2,1},{1,0,2},{1,2,0},{2,1,0},{2,0,1}};
inline int id(char c)
{
    if(c=='Q') return 0;
    if(c=='W') return 1;
    if(c=='E') return 2;
}
int ty[size];
inline int swap(char c)
{
    if(c=='Y') return 0;
    if(c=='V') return 1;
    if(c=='G') return 2;
    if(c=='C') return 3;
    if(c=='X') return 4;
    if(c=='Z') return 5;
    if(c=='T') return 6;
    if(c=='F') return 7;
    if(c=='D') return 8;
    if(c=='B') return 9;
}
int main()
{
    while(~scanf("%s",ss+1))
    {
        int len=strlen(ss+1);
        for(int i=0;i<=len;i++)
        {
            for(int j=0;j<=2;j++)
            {
                for(int k=0;k<=2;k++)
                {
                    dp[i][j][k]=inf;
                }
            }
        }
        for(int i=1;i<=len;i++)
        {
            ty[i]=swap(ss[i]);
            ans[i]=inf;
        }
        ans[1]=3;
        dp[1][id(s[ty[1]][0])][id(s[ty[1]][1])]=3;
        dp[1][id(s[ty[1]][0])][id(s[ty[1]][2])]=3;
        dp[1][id(s[ty[1]][1])][id(s[ty[1]][0])]=3;
        dp[1][id(s[ty[1]][1])][id(s[ty[1]][2])]=3;
        dp[1][id(s[ty[1]][2])][id(s[ty[1]][1])]=3;
        dp[1][id(s[ty[1]][2])][id(s[ty[1]][0])]=3;
        for(int i=2;i<=len;i++)
        {
            if(ty[i]==ty[i-1])
            {
                for(int j=0;j<=2;j++)
                {
                    for(int k=0;k<=2;k++)
                    {
                        dp[i][j][k]=dp[i-1][j][k];
                    }
                }
                ans[i]=ans[i-1];
                continue;
            }
            dp[i][id(s[ty[i]][0])][id(s[ty[i]][1])]=3+ans[i-1];
            dp[i][id(s[ty[i]][0])][id(s[ty[i]][2])]=3+ans[i-1];
            dp[i][id(s[ty[i]][1])][id(s[ty[i]][0])]=3+ans[i-1];
            dp[i][id(s[ty[i]][1])][id(s[ty[i]][2])]=3+ans[i-1];
            dp[i][id(s[ty[i]][2])][id(s[ty[i]][1])]=3+ans[i-1];
            dp[i][id(s[ty[i]][2])][id(s[ty[i]][0])]=3+ans[i-1];
            for(int j=0;j<6;j++)
            {
                dp[i][id(s[ty[i]][stru[j][1]])][id(s[ty[i]][stru[j][2]])]=min(dp[i][id(s[ty[i]][stru[j][1]])][id(s[ty[i]][stru[j][2]])],dp[i-1][id(s[ty[i]][stru[j][0]])][id(s[ty[i]][stru[j][1]])]+1);
                for(int k=0;k<=2;k++)
                dp[i][id(s[ty[i]][stru[j][1]])][id(s[ty[i]][stru[j][2]])]=min(dp[i][id(s[ty[i]][stru[j][1]])][id(s[ty[i]][stru[j][2]])],dp[i-1][k][id(s[ty[i]][stru[j][0]])]+2);
                ans[i]=min(ans[i],dp[i][id(s[ty[i]][stru[j][1]])][id(s[ty[i]][stru[j][2]])]);
            }
        }
        printf("%d\n",ans[len]+len);
    }
    return 0;
}