[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
5 1 5 2 2 1 3 2 1 2 1
Sample Output 1
6
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
3 1 1 1 3 2 1
Sample Output 2
0
All balls are in the same color, so no cost is incurred in swapping balls.
Sample Input 3
3 3 1 2 1 1 2
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);
}
