hdu 5900 QSC and Master 简单区间dp

最新推荐文章于 2020-04-28 22:01:05 发布

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一名学生在东北大学的东楼遇到了一位传说中的大师，并被要求解决一个数学难题：从成对出现的键值中选择连续的非互质键值对以获得最大得分。此问题与动态规划有关。

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Every school has some legends, Northeastern University is the same.

Enter from the north gate of Northeastern University，You are facing the main building of Northeastern University.Ninety-nine percent of the students have not been there，It is said that there is a monster in it.

QSCI am a curious NEU_ACMer，This is the story he told us.

It’s a certain period，QSCI am in a dark night, secretly sneaked into the East Building，hope to see the master.After a serious search，He finally saw the little master in a dark corner. The master said：

“You and I, we’re interfacing.please solve my little puzzle！

There are N pairs of numbers，Each pair consists of a key and a value，Now you need to move out some of the pairs to get the score.You can move out two continuous pairs，if and only if their keys are non coprime(their gcd is not one).The final score you get is the sum of all pair’s value which be moved out. May I ask how many points you can get the most？

The answer you give is directly related to your final exam results~The young man~”

QSC is very sad when he told the story，He failed his linear algebra that year because he didn’t work out the puzzle.

Could you solve this puzzle?

（Data range：1<=N<=300
1<=Ai.key<=1,000,000,000
0

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;
ll num[310];
ll dp[310][310];
struct node{
    int key, val;
}s[310];
int t[310][310];
int sum[310];
int gcd(int a,int b)
{
    if(b==0) return a;
    else return gcd(b,a%b);
}
int main()
{
    int N;
    cin>>N;
    while(N--)
    {
        int n;
        cin>>n;
        memset(dp,0,sizeof(dp));
        memset(t,0,sizeof(t));
        memset(num,0,sizeof(num));
        for(int i=0;i<n;i++)
            scanf("%d",&s[i].key);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&s[i].val);
            sum[i]=sum[i-1]+s[i].val;
        }
        for(int i=0;i<n;i++)
        {
            for(int j=i+1;j<n;j++)
                if(gcd(s[i].key,s[j].key)>1)
                    t[i][j]=1;
        }
        for(int i=0;i<n-1;i++)
        {
            if(t[i][i+1]) dp[i][i+1]=s[i].val+s[i+1].val;
        }
        for(int len=2;len<n;len++)
        {
            for(int i=0;i+len<n;i++)
            {
                int j=i+len;
                dp[i][j]=dp[i+1][j];
                for(int k=i+1;k<=j;k++)
                {
                    dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
                    if(t[i][k]&&dp[i+1][k-1]==sum[k-1]-sum[i])
                    dp[i][j]=max(dp[i][j],dp[i+1][k-1]+s[i].val+s[k].val+dp[k+1][j]);
                }
            }
        }
         printf("%lld\n",dp[0][n-1] );
    }
}