[ABC261F] Sorting Color Balls

Problem Statement

There are $N$ balls arranged from left to right.
The color of the $i$-th ball from the left is Color $C_i$, and an integer $X_i$ is written on it.

Takahashi wants to rearrange the balls so that the integers written on the balls are non-decreasing from left to right.
In other words, his objective is to reach a situation where, for every $1\leq i\leq N-1$, the number written on the $(i+1)$-th ball from the left is greater than or equal to the number written on the $i$-th ball from the left.

For this, Takahashi can repeat the following operation any number of times (possibly zero):

Choose an integer $i$ such that $1\leq i\leq N-1$.

If the colors of the $i$-th and $(i+1)$-th balls from the left are different, pay a cost of $1$.
(No cost is incurred if the colors are the same).

Swap the $i$-th and $(i+1)$-th balls from the left.

Find the minimum total cost Takahashi needs to pay to achieve his objective.

Constraints

$2 \leq N \leq 3\times 10^5$
$1\leq C_i\leq N$
$1\leq X_i\leq N$
All values in input are integers.

Input

Input is given from Standard Input in the following format:

$N$

$C_1$ $C_2$ $\ldots$ $C_N$

$X_1$ $X_2$ $\ldots$ $X_N$

Output

Print the minimum total cost Takahashi needs to pay to achieve his objective, as an integer.

Sample Input 1

Sample Output 1

Let us represent a ball as $($Color$,$ Integer$)$.
The initial situation is $(1,3)$, $(5,2)$, $(2,1)$, $(2,2)$, $(1,1)$.
Here is a possible sequence of operations for Takahashi:

Swap the $1$-st ball (Color $1$) and $2$-nd ball (Color $5$). Now the balls are arranged in the order $(5,2)$, $(1,3)$, $(2,1)$, $(2,2)$, $(1,1)$.
Swap the $2$-nd ball (Color $1$) and $3$-rd ball (Color $2$). Now the balls are arranged in the order $(5,2)$, $(2,1)$, $(1,3)$, $(2,2)$, $(1,1)$.
Swap the $3$-rd ball (Color $1$) and $4$-th ball (Color $2$). Now the balls are in the order $(5,2)$, $(2,1)$, $(2,2)$, $(1,3)$, $(1,1)$.
Swap the $4$-th ball (Color $1$) and $5$-th ball (Color $1$). Now the balls are in the order $(5,2)$, $(2,1)$, $(2,2)$, $(1,1)$, $(1,3)$.
Swap the $3$-rd ball (Color $2$) and $4$-th ball (Color $1$). Now the balls are in the order$(5,2)$, $(2,1)$, $(1,1)$, $(2,2)$, $(1,3)$.
Swap the $1$-st ball (Color $5$) and $2$-nd ball (Color $2$). Now the balls are in the order $(2,1)$, $(5,2)$, $(1,1)$, $(2,2)$, $(1,3)$.
Swap the $2$-nd ball (Color $5$) and $3$-rd ball (Color $1$). Now the balls are in the order $(2,1)$, $(1,1)$, $(5,2)$, $(2,2)$, $(1,3)$.

After the last operation, the numbers written on the balls are $1,1,2,2,3$ from left to right, which achieves Takahashi's objective.

The $1$-st, $2$-nd, $3$-rd, $5$-th, $6$-th, and $7$-th operations incur a cost of $1$ each, for a total of $6$, which is the minimum.
Note that the $4$-th operation does not incur a cost since the balls are both in Color $1$.

Sample Input 2

Sample Output 2

All balls are in the same color, so no cost is incurred in swapping balls.

Sample Input 3

首先当且仅当 $X_i<X_{i+1}$ 才会交换第 $i$ 个和第 $i+1$ 个。交换完后会减少一个逆序对。所以不考虑颜色，交换次数等于逆序对个数。

考虑颜色如果两个数颜色相同，那么不计价值。所以答案还要减去同颜色的逆序对个数即可。

#include<bits/stdc++.h>

using namespace std;

const int N=5005;

int x[N],c,y,t[N],n,m;

long long dp[N][N],ans;

int main()

{

	memset(dp,-0x7f,sizeof(dp));

	dp[0][0]=0;

	scanf("%d%d",&n,&m);

	for(int i=1;i<=n;i++)

		scanf("%d",x+i);

	for(int i=1;i<=m;i++)

		scanf("%d%d",&c,&y),t[c]+=y;

	for(int i=1;i<=n;i++)

	{

		dp[i][0]=dp[i-1][0];

		for(int j=1;j<=n;j++)

			dp[i][j]=dp[i-1][j-1]+t[j]+x[i],dp[i][0]=max(dp[i][0],dp[i-1][j-1]),ans=max(ans,dp[i][j]);

	}

	printf("%lld",ans);

}