Array Shrinking
題目描述
You are given an array a1,a2,…,an. You can perform the following operation any number of times:
Choose a pair of two neighboring equal elements ai=ai+1 (if there is at least one such pair).
Replace them by one element with value ai+1.
After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array a you can get?
Input
The first line contains the single integer n (1≤n≤500) — the initial length of the array a.
The second line contains n integers a1,a2,…,an (1≤ai≤1000) — the initial array a.
Output
Print the only integer — the minimum possible length you can get after performing the operation described above any number of times.
Examples
Input #1
5
4 3 2 2 3
Output #1
2
Input #2
7
3 3 4 4 4 3 3
Output #2
2
Input #3
3
1 3 5
Output #2
3
Input #4
1
1000
Output #4
1
Solution
區間dp.O（n^3）
令dp[i][j]表示從i到j區間進行合併後的最短長度。
dp[i][j] = dp[i][k] + dp[k + 1][j],
如果dp[i][k] 和 dp[k+1][j]都僅有一個數字且相同，那麼他們兩個可以合併，進行合併。
需要記錄一下i和j合併成一個數字後，該數字爲幾，comm[i][j]即爲i到j合併爲的一個數字。
轉移順序，區間dp的轉移方法，枚舉區間長度和左端點，算出右端點，枚舉斷點k。
答案爲dp[1][n]。
代碼
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int SZ = 500 + 20;
int n,dp[SZ][SZ],comm[SZ][SZ],num[SZ];
int main()
{
scanf("%d",&n);
for(int i = 1;i <= n;i ++ )
{
scanf("%d",&num[i]);
dp[i][i] = 1; //初始化
comm[i][i] = num[i];
}
for(int len = 2;len <= n;len ++ )
{
for(int i = 1;i <= n - len + 1;i ++ )
{
int j = i + len - 1;
dp[i][j] = INF;
for(int k = i;k <= j - 1;k ++ )
{
if(dp[i][j] >= dp[i][k] + dp[k + 1][j])
{
dp[i][j] = dp[i][k] + dp[k + 1][j]; //狀態轉移
if(dp[i][k] == 1 && dp[k + 1][j] == 1 && comm[i][k] == comm[k + 1][j]) //合併
{
dp[i][j] = 1;
comm[i][j] = comm[i][k] + 1;
}
}
}
}
}
printf("%d",dp[1][n]);
//system("pause");
return 0;
}
2020.3.17