HDU 5900 QSC and Master(区间dp)

来源：互联网发布：python genetic 编辑：程序博客网时间：2024/04/30 21:08

n≤300，然后相邻两个数字的key如果gcd！=1，就可以消除并且获得它的分数和
考虑区间dp[l][r]=max(dp[l+1][i−1]+b[l]+b[i]+dp[i+1][r]),gcd(a[l],a[i])!=1，s[l+1][i−1]=1
当然还得开个数组表示s[l][r]=1表示[l,r]之间可以全部消除
另外还有dp[l][r]=max(dp[l][r],dp[l+1][r])

代码：

#include <map>#include <set>#include <stack>#include <queue>#include <cmath>#include <string>#include <vector>#include <cstdio>#include <cctype>#include <cstring>#include <sstream>#include <cstdlib>#include <iostream>#include <algorithm>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;#define   MAX           1000005#define   MAXN          6005#define   maxnode       10#define   sigma_size    30#define   lson          l,m,rt<<1#define   rson          m+1,r,rt<<1|1#define   lrt           rt<<1#define   rrt           rt<<1|1#define   middle        int m=(r+l)>>1#define   LL            long long#define   ull           unsigned long long#define   mem(x,v)      memset(x,v,sizeof(x))#define   lowbit(x)     (x&-x)#define   pii           pair<int,int>#define   bits(a)       __builtin_popcount(a)#define   mk            make_pair#define   limit         10000//const int    prime = 999983;const int    INF   = 0x3f3f3f3f;const LL     INFF  = 0x3f3f;const double pi    = acos(-1.0);//const double inf   = 1e18;const double eps   = 1e-8;const LL     mod   = 1e9+7;const ull    mx    = 133333331;/*****************************************************/inline void RI(int &x) {    char c;    while((c=getchar())<'0' || c>'9');    x=c-'0';    while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0'; }/*****************************************************/int a[305];int b[305];LL dp[305][305];int s[305][305];LL dfs(int l,int r){    if(dp[l][r]!=-1) return dp[l][r];    if(l==r){        s[l][r]=-1;        dp[l][r]=0;        return dp[l][r];    }    if(l>r){        s[l][r]=1;        dp[l][r]=0;        return dp[l][r];    }    if(l+1==r){        if(__gcd(a[l],a[r])!=1){            s[l][r]=1;            dp[l][r]=b[l]+b[r];        }        else{            s[l][r]=-1;            dp[l][r]=0;        }        return dp[l][r];    }    LL ans=0;    int flag=0;    for(int i=l+1;i<=r;i++){        LL tmp=dfs(l+1,i-1);        if(s[l+1][i-1]==1&&__gcd(a[l],a[i])!=1){            ans=max(ans,dp[l+1][i-1]+b[l]+b[i]+dfs(i+1,r));            if(s[i+1][r]==1) flag=1;        }    }    if(flag) s[l][r]=1;    else s[l][r]=-1;    ans=max(ans,dfs(l+1,r));    return dp[l][r]=ans;}int main(){    int t;    cin>>t;    while(t--){        int n;        cin>>n;        for(int i=1;i<=n;i++) scanf("%d",&a[i]);        for(int i=1;i<=n;i++) scanf("%d",&b[i]);        mem(dp,-1);        cout<<dfs(1,n)<<endl;    }    return 0;}

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