hdu 4905 The Little Devil II 区间DP 四边形不等式优化

最新推荐文章于 2021-03-06 19:46:39 发布

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本文探讨了在有限步骤内优化数值序列的问题，通过选择适当的序列操作，最终达到最大分数的目标。介绍了四边形不等式优化技术在解决此类问题中的应用，并提供了具体的算法实现。在击败邪恶势力后，不仅拯救了国家，还揭示了解题背后的数学之美。

The Little Devil II

Time Limit: 16000/8000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 307 Accepted Submission(s): 78

Problem Description

There is an old country and the king fell in love with a devil. The devil always asks the king to do some crazy things. Although the king used to be wise and beloved by his people. Now he is just like a boy in love and can’t refuse any request from the devil. Also, this devil is looking like a very cute Loli.

Y*wan defeat the princess's knight party, and finally he meet the devil. But, after that, y*wan feel in love with the devil and become a lolicon. As the king is already died, y*wan become the new king.

Some days passed, the devil feel bored about this boring country and leaves, but the lolicon remains in the country.

The finally hero, you comes as promised, and you figure out a hard problem to defeat y*wan:

There is n integers a_1,a_2,...,a_n on a line, each time we can take two adjacent integers a and b, and replace them by their gcd(a,b), and add gcd(a,b) to the total score. The score is initially the sum of all a_i.After n-1 steps there is only one number left and you stop. What is your maximum score in the end?

After defeat y*wan, you kill WJMZBMR, and save the country.

Input

The first line contains an integer T, denoting the number of the test cases.
For each test case, the first line contains an integer n.
The next line contains n integers a_1,a_2,...,a_n separated by a single space.

T<=20, n<=3000.
a_i >=0
a_i is in the range of int(C++).

Output

For each test case, output the result in one line.

Sample Input

Sample Output

多校第四场，比赛的时候超时。。

可以把任意两个相邻的数变成他们的gcd，并且把这个gcd加到最后的答案中去。选择合适的顺序，使最后的答案最大。

好吧。。今天才知道有四边形不等式优化这个东西。。

这里介绍的很详细点击打开链接

总之如果满足四边形不等式的条件的话，区间dp的时候要记录s[i][j]，第三重循环的范围是s[i][j-1]-s[i+1][j]。

#include<iostream>
#include<algorithm>
#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#define INF 0x3f3f3f3f
#define MAXN 3010
#define MAXM 110
#define MOD 1000000007
#define MAXNODE 4*MAXN
#define eps 1e-9
using namespace std;
typedef long long LL;
int T,N;
int a[MAXN],g[MAXN][MAXN],s[MAXN][MAXN];
LL dp[MAXN][MAXN];
int gcd(int a,int b){
    return a%b?gcd(b,a%b):b;
}
void init(){
    for(int i=1;i<=N;i++){
        g[i][i]=a[i];
        for(int j=i+1;j<=N;j++) g[i][j]=gcd(g[i][j-1],a[j]);
    }
}
int main(){
    freopen("in.txt","r",stdin);
    scanf("%d",&T);
    while(T--){
        scanf("%d",&N);
        for(int i=1;i<=N;i++) scanf("%d",&a[i]);
        init();
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=N;i++){
            s[i][i]=i;
            dp[i][i]=a[i];
        }
        for(int l=1;l<N;l++)
            for(int i=1;i+l<=N;i++){
                int j=i+l;
                for(int k=s[i][j-1];k<=s[i+1][j];k++){
                    if(k<j&&dp[i][j]<dp[i][k]+dp[k+1][j]+g[i][j]){
                        dp[i][j]=dp[i][k]+dp[k+1][j]+g[i][j];
                        s[i][j]=k;
                    }
                }
            }
        printf("%I64d\n",dp[1][N]);
    }
    return 0;
}