2019ccpc秦皇島I Invoker

題意：10種技能，每種技能需要集齊3個給定的字符（無序）才能召喚。英雄每個時刻都只能擁有3個字符，並且每發動一次技能需要一個額外的字符。輸入一個技能字符串，求依次發動它們的最小花費字符數。
思路：10種技能，每種技能無序3！==6種組合，打表打出來，據此線性dp。
寫了代碼，還沒有交，存一下。

#include<bits/stdc++.h>
using namespace std;

char s[10][6][4]={
    {"QQQ","QQQ","QQQ","QQQ","QQQ","QQQ"},
    {"QQW","QWQ","WQQ","QQW","QWQ","WQQ"},
    {"QQE","QEQ","EQQ","QQE","QEQ","EQQ"},
    {"WWW","WWW","WWW","WWW","WWW","WWW"},
    {"QWW","WQW","WWQ","QWW","WQW","WWQ"},
    {"WWE","WEW","EWW","WWE","WEW","EWW"},
    {"EEE","EEE","EEE","EEE","EEE","EEE"},
    {"QEE","EQE","EEQ","QEE","EQE","EEQ"},
    {"WEE","EWE","EEW","WEE","EWE","EEW"},
    {"QWE","QEW","WQE","WEQ","EWQ","EQW"}
};
char t[11]="YVGCXZTFDB";
char str[100000+100];
int f[100000+100][6];
map<char,int> mp;

int main()
{
    //freopen("input.in","r",stdin);
    for(int i=0;i<10;i++)mp[t[i]]=i;
    while(scanf("%s",str)!=EOF)
    {
        int len=strlen(str);
        for(int j=0;j<6;j++)f[0][j]=3;
        for(int i=1;i<len;i++)
        {
            int last=mp[str[i-1]],now=mp[str[i]];
            if(last==now)
            {
                for(int j=0;j<6;j++)f[i][j]=f[i-1][j];
                continue;
            }
            for(int j=0;j<6;j++)
            {
                f[i][j]=(1<<30);
                for(int k=0;k<6;k++)
                {
                    if(s[last][k][1]==s[now][j][0] && s[last][k][2]==s[now][j][1])f[i][j]=min(f[i][j],f[i-1][k]+1);
                    else if(s[last][k][2]==s[now][j][0])f[i][j]=min(f[i][j],f[i-1][k]+2);
                    else f[i][j]=min(f[i][j],f[i-1][k]+3);
                }
            }
        }
        int minn=(1<<30);
        for(int j=0;j<6;j++)minn=min(minn,f[len-1][j]);
        minn+=len;
        cout<<minn<<endl;
    }   
    return 0;
}