2019ccpc秦皇岛/Gym102361 I - Invoker dp

题意：

连续3个特定的按键（在这3个中不要求顺序）能使出某个技能，使出不同技能所需要的按键可以重叠，给你一个技能序列，问你最少花费多少次按键能按顺序使出这些招数。

题解：

dp，dp[i][j]代表使出第i个技能后，最后3个按键组合是j的最少需要按键数，暴力枚举即可。

思路比较绕，实际码的时候很自闭。

#include<bits/stdc++.h>
using namespace std;
#define MAXN 100000+50
char s[MAXN];
string jineng[][]={
    {"QQQ","QQQ","QQQ","QQQ","QQQ","QQQ"},
    {"QQW","QQW","QWQ","QWQ","WQQ","WQQ"},
    {"QQE","QQE","QEQ","QEQ","EQQ","EQQ"},
    {"WWW","WWW","WWW","WWW","WWW","WWW"},
    {"QWW","QWW","WQW","WQW","WWQ","WWQ"},
    {"WWE","WWE","WEW","WEW","EWW","EWW"},
    {"EEE","EEE","EEE","EEE","EEE","EEE"},
    {"QEE","QEE","EQE","EQE","EEQ","EEQ"},
    {"WEE","WEE","EEW","EEW","EWE","EWE"},
    {"QWE","QEW","WEQ","WQE","EWQ","EQW"}
};
//CXZTFDB
int num[];
int dp[MAXN][];
int get_dif(int id1,int id2,int pos1,int pos2)
{
    if(jineng[id1][pos1][]==jineng[id2][pos2][]&&jineng[id1][pos1][]==jineng[id2][pos2][]
    &&jineng[id1][pos1][]==jineng[id2][pos2][])
    return ;
    if(jineng[id1][pos1][]==jineng[id2][pos2][]&&jineng[id1][pos1][]==jineng[id2][pos2][])
    return ;
    if(jineng[id1][pos1][]==jineng[id2][pos2][]) return ;
    return ;
}
int main()
{
    int ans;
    scanf("%s",s);
    int len=strlen(s);
    ans=len*;
    memset(num,,sizeof(num));
    num['Y']=,num['V']=,num['G']=,num['C']=,
    num['X']=,num['Z']=,num['T']=,num['F']=,
    num['D']=,num['B']=;
    for(int i=;i<len;++i)
        for(int j=;j<;++j)
            dp[i][j]=len*;
    for(int i=;i<;++i) dp[][i]=;
    for(int i=;i<len;++i)
        for(int j=;j<;++j)
            for(int k=;k<;++k)
            {
                dp[i][j]=min(dp[i][j],dp[i-][k]+get_dif(num[s[i-]],num[s[i]],k,j));
            }
    for(int i=;i<;++i) ans=min(ans,dp[len-][i]);
    cout<<ans+len<<endl;
}