区间DPB. Zuma

Posted 2022-04-14 行码棋

https://codeforces.com/problemset/problem/607/B

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$f[i][j]$：代表区间$[i,j]$合并的最小花费

• 初始化：
  $f[i][i]=1$
  $f[i][i+1]=1,a[i]==a[i+1]$
  $f[i][i+1]=2,a[i]\ne a[i+1]$
• 转移方程：
  $f[i][j]=f[i+1][j-1],a[i]==a[j]$
  $f[i][j]=min(f[i][k]+f[k+1][j]),i\le k<j$

区间DP一般都是由两个小的区间合并成一个大的区间，所以第三维循环非常常见，应该对此非常敏感

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

const int N = 505, mod = 19650827;
int f[N][N], a[N];

int main()

	ios::sync_with_stdio(false);
	cin.tie(0);
	
	int n;
	cin >> n;
	for(int i = 1; i <= n; i++)
		cin >> a[i];
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n; j++)
			f[i][j] = 0x3f3f3f3f;
	for(int i = 1; i <= n; i++)
	
		f[i][i] = 1;
		if(i < n) f[i][i + 1] = 1 + (a[i] != a[i + 1]);
	
	for(int len = 3; len <= n; len ++)
	
		for(int i = 1; i + len - 1 <= n; i++)
		
			int j = i + len - 1;
			if(a[i] == a[j])
				f[i][j] = f[i + 1][j - 1];
			for(int k = i; k < j; k++)
				f[i][j] = min(f[i][j], f[i][k] + f[k + 1][j]);
		
	
	cout << f[1][n] << "\\n";
	return 0;

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