CF665B Shopping

CF665B Shopping

题目描述

Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online.

The store contains $ k $ items. $ n $ customers have already used the above service. Each user paid for $ m $ items. Let $ a_{ij} $ denote the $ j $ -th item in the $ i $ -th person’s order.

Due to the space limitations all the items are arranged in one single row. When Ayush receives the $ i $ -th order he will find one by one all the items $ a_{ij} $ ( $ 1<=j<=m $ ) in the row. Let $ pos(x) $ denote the position of the item $ x $ in the row at the moment of its collection. Then Ayush takes time equal to $ pos(a_{i1})+pos(a_{i2})+…+pos(a_{im}) $ for the $ i $ -th customer.

When Ayush accesses the $ x $ -th element he keeps a new stock in the front of the row and takes away the $ x $ -th element. Thus the values are updating.

Your task is to calculate the total time it takes for Ayush to process all the orders.

You can assume that the market has endless stock.

输入格式

The first line contains three integers

n

n

n ,

m

m

m and

k

k

k (

1

≤

n

,

k

≤

100

,

1

≤

m

≤

k

1\le n,k\le 100,1\le m\le k

1≤n,k≤100,1≤m≤k ) — the number of users, the number of items each user wants to buy and the total number of items at the market.

The next line contains

k

k

k distinct integers

p

l

p_{l}

pl​ (

1

<

=

p

l

<

=

k

1<=p_{l}<=k

1<=pl​<=k ) denoting the initial positions of the items in the store. The items are numbered with integers from

1

1

1 to

k

k

k .

Each of the next

n

n

n lines contains

m

m

m distinct integers

a

i

j

a_{ij}

aij​ (

1

<

=

a

i

j

<

=

k

1<=a_{ij}<=k

1<=aij​<=k ) — the order of the

i

i

i -th person.

输出格式

Print the only integer $ t $ — the total time needed for Ayush to process all the orders.

输入输出样例

样例输入1

2 2 5
3 4 1 2 5
1 5
3 1

样例输出1

14

说明/提示

Customer

1

1

1 wants the items

1

1

1 and

5

5

5 .

p

o

s

(

1

)

=

3

pos(1)=3

pos(1)=3 , so the new positions are:

[

1

,

3

,

4

,

2

,

5

]

[1,3,4,2,5]

[1,3,4,2,5] .

p

o

s

(

5

)

=

5

pos(5)=5

pos(5)=5 , so the new positions are:

[

5

,

1

,

3

,

4

,

2

]

[5,1,3,4,2]

[5,1,3,4,2] .

Time taken for the first customer is

3

+

5

=

8

3+5=8

3+5=8 .

Customer

2

2

2 wants the items

3

3

3 and

1

1

1 .

p

o

s

(

3

)

=

3

pos(3)=3

pos(3)=3 , so the new positions are:

[

3

,

5

,

1

,

4

,

2

]

[3,5,1,4,2]

[3,5,1,4,2] .

p

o

s

(

1

)

=

3

pos(1)=3

pos(1)=3 , so the new positions are:

[

1

,

3

,

5

,

4

,

2

]

[1,3,5,4,2]

[1,3,5,4,2] .

Time taken for the second customer is

3

+

3

=

6

3+3=6

3+3=6 .

Total time is

8

+

6

=

14

8+6=14

8+6=14 .

Formally

p

o

s

(

x

)

pos(x)

pos(x) is the index of

x

x

x in the current row.

思路

题上首先给了三个数

n

,

m

,

k

n,m,k

n,m,k，然后在第二行里给出了

k

k

k 个数，然后又

n

×

m

n\times m

n×m 个操作，代表每一次把这个操作的数移动到第一位，其他的往后推，每次移动的时候有一个权值（从第一个数到第这个数的相差的位置数），问的是所有权值的和，数据量不大，直接模拟就好。

注

看到楼上用的 vector，我这里用数组。

Code

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

int main()
{
	int n, m, k, x, ans = 0, a[105];
	scanf("%d %d %d", &n, &m, &k);
	for (register int i(1); i <= k; ++i)
		scanf("%d", &a[i]);
	for (register int i(1); i <= n * m; ++i)
	{
		scanf("%d", &x);
		register int j, l;
		for (j = 1; j <= k; ++j)
			if (a[j] == x)
				break;
		ans += j;
		for (l = j; l >= 2; --l)
			a[l] = a[l - 1];
		a[1] = x;
	}
	printf("%d\n", ans);
	return 0;
}

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