CF 1114D-Flood Fill-n2复杂度的区间dp

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#CF 1114D #Flood Fill #区间dp

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本文介绍了一种解决Flood Fill游戏的算法，通过区间动态规划在O(n^2)的时间复杂度下，找到将一串彩色方块统一为单一颜色所需的最少操作次数。该算法适用于方块数量不超过5000的情况，通过比较和更新不同颜色段的转换成本，最终得出全局最优解。

Description

You are given a line of n colored squares in a row, numbered from 1 to n from left to right. The i-th square initially has the color ci.
Let’s say, that two squares i and j belong to the same connected component if ci=cj, and ci=ck for all k satisfying i<k<j. In other words, all squares on the segment from i to j should have the same color.
For example, the line [3,3,3] has 1 connected component, while the line [5,2,4,4] has 3 connected components.
The game “flood fill” is played on the given line as follows:

At the start of the game you pick any starting square (this is not counted as a turn).
Then, in each game turn, change the color of the connected component containing the starting square to any other color.

Find the minimum number of turns needed for the entire line to be changed into a single color.

Input

The first line contains a single integer n (1≤n≤5000) — the number of squares.
The second line contains integers c1,c2,…,cn (1≤ci≤5000) — the initial colors of the squares.

Output

Print a single integer — the minimum number of the turns needed.

Sample Input

8
4 5 2 2 1 3 5 5

Sample Output

核心思想：

O(n²)的复杂度写区间dp。

代码如下：

#include<cstdio>
#include<iostream>
using namespace std;
typedef long long ll;
const int N=5e3+5;
int dp[N][N],a[N];
int main()
{
	int n;
	scanf("%d",&n);
	n=2;
	scanf("%d",&a[1]);
	//压缩连续相等的值 
	while(scanf("%d",&a[n])!=EOF)
		if(a[n]!=a[n-1])
			n++;
	n--;
	for(int d=2;d<=n;d++)
		for(int i=1;i+d-1<=n;i++)
		{
			int j=i+d-1;
			//相等则中间区间+1 
			if(a[i]==a[j])
			{
				if(d==2)
					dp[i][j]=0;
				else
					dp[i][j]=dp[i+1][j-1]+1;
			}
			else
			{
				//源点只有一个，来自左区间或右区间 
				dp[i][j]=min(dp[i][j-1],dp[i+1][j])+1;
			}
		}
	printf("%d\n",dp[1][n]);
	return 0;
}