Invoker(小DP）

题意：http://acm.hdu.edu.cn/showproblem.php?pid=6739

尽量让他们连起来。

思路：

直接dp，其中一个状态是以什么结尾。

#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
#include <cstdio>//sprintf islower isupper
#include <cstdlib>//malloc  exit strcat itoa system("cls")
#include <iostream>//pair
#include <fstream>//freopen("C:\Users\13606\Desktop\草稿.txt","r",stdin);
#include <bitset>
//#include <map>
//#include<unordered_map>
#include <vector>
#include <stack>
#include <set>
#include <string.h>//strstr substr
#include <string>
#include <time.h>//srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
#include <cmath>
#include <deque>
#include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
#include <vector>//emplace_back
//#include <math.h>
//#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
#include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
//******************
int abss(int a);
int lowbit(int n);
int Del_bit_1(int n);
int maxx(int a,int b);
int minn(int a,int b);
double fabss(double a);
void swapp(int &a,int &b);
clock_t __STRAT,__END;
double __TOTALTIME;
void _MS(){__STRAT=clock();}
void _ME(){__END=clock();__TOTALTIME=(double)(__END-__STRAT)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
//***********************
#define rint register int
#define fo(a,b,c) for(rint a=b;a<=c;++a)
#define fr(a,b,c) for(rint a=b;a>=c;--a)
#define mem(a,b) memset(a,b,sizeof(a))
#define pr printf
#define sc scanf
#define ls rt<<1
#define rs rt<<1|1
typedef long long ll;
const double E=2.718281828;
const double PI=acos(-1.0);
//const ll INF=(1LL<<60);
const int inf=(1<<30);
const double ESP=1e-9;
const int mod=(int)1e9+7;
const int N=(int)1e5+10;

int dp[N][50];
char s[N];
int tot;
int ID[400];
int map[50];
int id(int x)
{
    return ID[x];
}
struct node
{
    int a,b,c;
}p[10];
void PR(int _[],int n)
{
    for(int i=1;i<=n;++i)
        pr("%d ",_[i]);
    pr("
");
}
int main()
{
    tot=0;
    for(int i=1;i<=350;++i)
    {
        int temp=i;
        bool f=1;
        while(temp>0)
        {
            if(temp%10!=1&&temp%10!=2&&temp%10!=3)
                f=0;
            temp/=10;
        }
        if(f)map[++tot]=i;
    }
    for(int i=1;i<=45;++i)
        ID[map[i]]=i;
    p[1]={1,2,3};
    p[2]={1,3,2};
    p[3]={2,1,3};
    p[4]={2,3,1};
    p[5]={3,1,2};
    p[6]={3,2,1};
    int v[200];
    v['Y']=111;
    v['V']=112;
    v['G']=113;
    v['C']=222;
    v['X']=122;
    v['Z']=223;
    v['T']=333;
    v['F']=133;
    v['D']=233;
    v['B']=123;
    while(~sc("%s",s+1))
    {
        int l=strlen(s+1);
        for(int i=1;i<=l;++i)
            for(int j=1;j<=45;++j)
                dp[i][j]=inf;
        for(int i=1;i<=6;++i)
        {
            int temp=v[s[1]];
            int cnt=3;
            int mp[5];
            while(temp>0)
            {
                mp[cnt]=temp%10;
                temp/=10;
                cnt--;
            }
            int x=mp[p[i].b]*10+mp[p[i].c];
            dp[1][id(mp[p[i].a]*100+mp[p[i].b]*10+mp[p[i].c])]=min(3,dp[1][id(mp[p[i].a]*100+mp[p[i].b]*10+mp[p[i].c])]);
            dp[1][id(x)]=min(dp[1][id(x)],3);
            x=mp[p[i].c];
            dp[1][id(x)]=min(dp[1][id(x)],3);
        }
        //PR(dp[1],37);
        int ans=0;
        for(int i=2;i<=l+1;++i)
        {
            int min_=inf;
            for(int j=1;j<=45;++j)
                min_=min(min_,dp[i-1][j]);
            ans=min_;
            if(i==l+1)break;

            int temp=v[s[i]];
            int cnt=3;
            int mp[5];
            while(temp>0)
            {
                mp[cnt]=temp%10;
                temp/=10;
                cnt--;
            }
            for(int j=1;j<=6;++j)
            {
                int x=mp[p[j].a],y=mp[p[j].b],z=mp[p[j].c];
            //    cout<<x<<' '<<y<<' '<<z<<endl;
                dp[i][id(x*100+y*10+z)]=min(dp[i][id(x*100+y*10+z)],min(min(dp[i-1][id(x)]+2,min(min_+3,dp[i-1][id(x*10+y)]+1)),dp[i-1][id(x*100+y*10+z)]));
                dp[i][id(y*10+z)]=min(dp[i][id(y*10+z)],min(min(dp[i-1][id(x)]+2,min(min_+3,dp[i-1][id(x*10+y)]+1)),dp[i-1][id(x*100+y*10+z)]));
                dp[i][id(z)]=min(dp[i][id(z)],min(min(dp[i-1][id(x)]+2,min(min_+3,dp[i-1][id(x*10+y)]+1)),dp[i-1][id(x*100+y*10+z)]));
            }
        //    PR(dp[i],37);
        }
        pr("%d
",ans+l);
    }
    return 0;
}

/**************************************************************************************/

int maxx(int a,int b)
{
    return a>b?a:b;
}

void swapp(int &a,int &b)
{
    a^=b^=a^=b;
}

int lowbit(int n)
{
    return n&(-n);
}

int Del_bit_1(int n)
{
    return n&(n-1);
}

int abss(int a)
{
    return a>0?a:-a;
}

double fabss(double a)
{
    return a>0?a:-a;
}

int minn(int a,int b)
{
    return a<b?a:b;
}