2016 ACM/ICPC Asia Regional Shenyang Online HDU 5900 QSC and Master(区间dp水题)

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本文介绍了一道编程竞赛题目，通过区间动态规划解决非互质数对的最大得分问题，给出AC代码及解析。

QSC and Master

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 663 Accepted Submission(s): 240

Problem Description

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<Ai.value<=1,000,000,000）

Input

First line contains a integer T，means there are T(1≤T≤10) test case。

Each test case start with one integer N . Next line contains N integers，means Ai.key.Next line contains N integers，means Ai.value.

Output

For each test case,output the max score you could get in a line.

Sample Input

Sample Output

0
2
0

【题意】n 个pair<int , int>，每次可以选相邻两个pair。如果他们的first不互质就可以把它们都删掉，并且获得second之和的分数，问最大得分。
【解题方法】区间dp水题。定义dp[i][j] 表示i~j的区间能得的最大分，那么有状态转移方程：dp[i][j]=dp[i][k]+dp[k+1][j]。
【AC 代码】

//
//Created by just_sort 2016/9/12 17:01
//Copyright (c) 2016 just_sort.All Rights Reserved
//

#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 305;
const int maxm = 305;
LL dp[maxn][maxn];
LL a[maxn],b[maxn],sum[maxn];

int main()
{
    int T,n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1; i<=n; i++){
            scanf("%I64d",&a[i]);
        }
        sum[0]=0LL;
        for(int i=1; i<=n; i++){
            scanf("%I64d",&b[i]);
            sum[i]=sum[i-1]+b[i];
        }
        memset(dp,0,sizeof(dp));
        for(int len=1; len<=n; len++){
            for(int i=1; i+len<=n; i++){
                if(dp[i+1][i+len-1]==sum[i+len-1]-sum[i]){
                    dp[i][i+len] = dp[i+1][i+len-1];
                    if(__gcd(a[i],a[i+len])>1) dp[i][i+len] += b[i]+b[i+len];
                }
                for(int k=i; k<i+len; k++){
                    dp[i][i+len] = max(dp[i][i+len],dp[i][k]+dp[k+1][i+len]);
                }
            }
        }
        printf("%I64d\n",dp[1][n]);
    }
    return 0;
}