HDU 5900 QSC and Master 区间dp

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5900

QSC and Master

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 131072/131072 K (Java/Others) #### 问题描述 > 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输入

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.

输出

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

样例输入

3
3
1 2 3
1 1 1
3
1 2 4
1 1 1
4
1 3 4 3
1 1 1 1

样例输出

0
2
0

题意

给你若干个数，如果相邻的两个数不互质，那么就可以吧它们消掉，并且取得这两个数的权值，问如何消能过获得最大的权值和。

题解

区间dp，dp[i][j]表示区间[i,j]能够获得的最大权值的方案，这里有个贪心策略，如果能全部消掉，我们肯定选择全部消。

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef __int64 LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=333;

LL arr[maxn],sumv[maxn],val[maxn];
LL dp[maxn][maxn];

LL gcd(LL a,LL b){
    return b==0?a:gcd(b,a%b);
}

LL dfs(int l,int r){
    if(dp[l][r]>=0) return dp[l][r];
    if(l==r) return dp[l][r]=0;
    if(l==r-1){
        if(gcd(arr[l],arr[r])==1) return dp[l][r]=0;
        else return dp[l][r]=val[l]+val[r];
    }
    LL &res=dp[l][r]=0;
    for(int k=l;k<r;k++){
        res=max(res,dfs(l,k)+dfs(k+1,r));
    }
    LL tmp=dfs(l+1,r-1);
    if(tmp==sumv[r-1]-sumv[l]&&gcd(arr[l],arr[r])!=1) res=max(res,tmp+val[l]+val[r]);
    else res=max(res,tmp);

    return res;
}

void init(){
    clr(dp,-1);
}

int n;

int main() {
    int tc,kase=0;
    scf("%d",&tc);
    while(tc--){
        init();
        scf("%d",&n);
        for(int i=1;i<=n;i++) scf("%I64d",&arr[i]);
        for(int i=1;i<=n;i++) scf("%I64d",&val[i]);
        sumv[0]=0;
        for(int i=1;i<=n;i++) sumv[i]=sumv[i-1]+val[i];
        prf("%I64d\n",dfs(1,n));
    }
    return 0;
}

//end-----------------------------------------------------------------------