[原创]codeforces 607B #336B Zuma [区间DP]【动态规划】

2017-01-12 14:25:19 Tabris_ 阅读数：552

博客爬取于2020-06-14 22:42:15
以下为正文

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https://blog.csdn.net/qq_33184171/article/details/54377643

题目连接:http://codeforces.com/contest/607/problem/B
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B. Zuma
time limit per test2 seconds
memory limit per test512 megabytes
inputstandard input
outputstandard output
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.
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题目大意：
就是在长度为n的序列中，每次能够消去任意一段回文字串，问最少多少次才能使这个序列全部消去

解题思路：
区间DP
dp[i][j] 表示消去区间 $[i,j]$ 的最小次数

那么首先对于回文的时候也就是
$a_i==a_j$ 的时候 dp[i][j]=dp[i+1][j-1];

对于其他情况我们可以枚举断点，计算区间dp[i][j]的最优解.即:
dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]);

写的时候用得记忆化搜索的形式,相比for循环会少思考dp的先后关系,能减少一些错误.

附本题代码
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# include <bits/stdc++.h>
using namespace std;
# define abs(x) ((x>=0)?(x):-(x))
typedef long long int LL;
const int MOD = 1e9;
const int N = 1e6+7;
/*******************************************/
int solve(int l,int r){
    if(l>=r) return 1;
    if(dp[l][r]!=-1) return dp[l][r];
    dp[l][r]=1000000;
    if(a[l]==a[r]) dp[l][r]=solve(l+1,r-1);
    for(int i=l;i<r;i++) dp[l][r]=min(dp[l][r],solve(l,i)+solve(i+1,r));
    return dp[l][r];
}

int main(){

    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    for(int i=0;i<=n+1;++i)for(int j=0;j<=n+1;++j) dp[i][j]=-1;
    printf("%d\n",solve(1,n));
    return 0;
}