Prediction and Restriction——UPC

题目描述

At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
·The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
·Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
→R points for winning with Rock;
→S points for winning with Scissors;
→P points for winning with Paper.
·However, in the i-th round, the player cannot use the hand he/she used in the (i−K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1≤i≤N)
is r, the machine will play Rock in the i-th round. Similarly, p and s stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?

Notes
In this problem, Rock Paper Scissors can be thought of as a two-player game, in which each player simultaneously forms Rock, Paper, or Scissors with a hand.
·If a player chooses Rock and the other chooses Scissors, the player choosing Rock wins;
·if a player chooses Scissors and the other chooses Paper, the player choosing Scissors wins;
·if a player chooses Paper and the other chooses Rock, the player choosing Paper wins;
·if both players play the same hand, it is a draw.

Constraints
·2≤N≤10⁵
·1≤K≤N−1
·1≤R,S,P≤10⁴
·N,K,R,S, and P are all integers.
·|T|=N
·T consists of r, p, and s.

输入

Input is given from Standard Input in the following format:

N K
R S P
T

输出

Print the maximum total score earned in the game.

样例输入

【样例1】
5 2
8 7 6
rsrpr
【样例2】
7 1
100 10 1
ssssppr
【样例3】
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr

样例输出

【样例1】
27
【样例2】
211
【样例3】
4996

提示

样例1解释
The machine will play {Rock, Scissors, Rock, Paper, Rock}.
We can, for example, play {Paper, Rock, Rock, Scissors, Paper} against it to earn 27 points. We cannot earn more points, so the answer is 27.

#pragma GCC optimize (2)
#pragma G++ optimize (2)
#include <bits/stdc++.h>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define wuyt main
typedef long long ll;
#define HEAP(...) priority_queue<__VA_ARGS__ >
#define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > >
template<class T> inline T min(T &x,const T &y){return x>y?y:x;}
template<class T> inline T max(T &x,const T &y){return x<y?y:x;}
//#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();
if(c == '-')Nig = -1,c = getchar();
while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();
return Nig*x;}
#define read read()
const ll inf = 1e15;
const int maxn = 2e5 + 7;
const int mod = 1e9 + 7;
#define start int wuyt()
#define end return 0
int n, r, s, p, k;
int judge[1005];///ASCII从97--122其实开200就已经够了
char pr[maxn], gr[maxn];
start{
    /**
    ll n=read,m=read;
    char judge[10];
    ll cnt=0;
    while(m--){
        cin>>judge+1;
        if(judge[3]=='y'){
            if(cnt==0) printf("F
");
            else if(cnt!=0) printf("T
");
        }
        else if(judge[3]=='u') printf("%lld
",cnt);
        else if(judge[3]=='n'){
            ll shu=read;
            if(num[shu]==0) printf("F
");
            if(num[shu]==1) printf("T
");
        }
        else if(judge[3]=='l'){
            ll temp=read;
            if(num[temp]==0){
                num[temp]=1;
                cnt++;
            }
            else if(num[temp]==1){
                num[temp]=0;
                cnt--;
            }
        }
    }
    **/
    n=read,k=read,r=read,s=read,p=read;
    judge['r']=r;
    judge['s']=s;
    judge['p']=p;
    cin>>pr+1;
    for (int i=1;i<=n;i++) {
        if (pr[i]=='s') pr[i]=gr[i]='r';
        else if (pr[i]=='p') pr[i] = gr[i] = 's';
        else pr[i]=gr[i]='p';
    }
    for (int i=1; i<=n;i++)
        if(i+k<=n&&gr[i]==gr[i+k])
            gr[i+k]='9';///重复
    int ans=0;
    for(int i=1;i<=n;i++)
        if(pr[i]==gr[i])
            ans+=judge[gr[i]];
    printf("%d
", ans);
    end;
}
 
/**************************************************************
    Language: C++
    Result: 正确
    Time:11 ms
    Memory:2416 kb
****************************************************************/