codeforces608D(85/600)

最新推荐文章于 2025-06-11 21:58:25 发布

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本文探讨了Zuma游戏中的一个算法问题：如何用最少的步骤消除所有宝石。通过动态规划的方法，实现了一个时间复杂度为O(n^3)的解决方案，并提供了完整的代码示例。

Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th of which has color ci. The goal of the game is to destroy all the gemstones in the line as quickly as possible.

In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?

Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.

Input
The first line of input contains a single integer n (1 ≤ n ≤ 500) — the number of gemstones.

The second line contains n space-separated integers, the i-th of which is ci (1 ≤ ci ≤ n) — the color of the i-th gemstone in a line.

Output
Print a single integer — the minimum number of seconds needed to destroy the entire line.

Examples
input
3
1 2 1
output
1
input
3
1 2 3
output
3
input
7
1 4 4 2 3 2 1
output
2
Note
In the first sample, Genos can destroy the entire line in one second.

In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.

In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.

上来就是一个三次方暴力

#include<bits/stdc++.h>
using namespace std;
int dp[501][501],tu[501];
int main()
{
    memset(dp,0x3f,sizeof(dp));
    int n;
    cin>>n;
    for(int a=1;a<=n;a++)cin>>tu[a];
    for(int a=1;a<=n;a++)dp[a][a]=1;
    for(int a=1;a<=n;a++)
    {
        for(int b=a-1;b>=1;b--)
        {
            if(tu[a]==tu[b])
            {
                if(a==b+1)
                {
                    dp[b][a]=1;
                    continue;
                }
                dp[b][a]=min(dp[b][a],dp[b+1][a-1]);
        //      dp[b][a]=min(dp[b][a],dp[b+1][a]+1);
        //      dp[b][a]=min(dp[b][a],dp[b][a-1]+1);
            }
            for(int c=b;c<=a;c++)
            {
                dp[b][a]=min(dp[b][a],dp[b][c]+dp[c+1][a]);
            }
        }
    }
//  cout<<dp[2][3]<<endl;
    cout<<dp[1][n]<<endl;
}