[GCD][区间DP]JZOJ 6287 扭动的树

escription

 

Input

Output

 

Sample Input

Sample 1:
4
2 3
6 4
9 8
12 1

Sample 2:
20
64978574415886122 263411
40589037247202745 239844
19724737874528206 167360
49216095485959384 760606
65063121727264647 659450
16572376111094320 726552
72014092598616298 133699
52843699826658793 427487
43374492289647376 552030
22047612465142862 605387
92386136280598953 718860
6436388687842008 368771
87727847161227820 880866
43622103777719758 352810
36870904328895185 322737
48993192459657624 456880
93250693206986868 619976
77407991580158822 861256
974508361120026 344635
77136053229840400 465474

Sample Output

Sample1:
51

Sample2:
101007480

 

Data Constraint

分析

比较显然的是，某个子树一定是一段键值连续的区间

所以就是个区间DP，合并时判断是否满足GCD条件即可

 

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
const ll Inf=1ll<<62;
const int N=310;
struct Node {
    ll k,v;
}a[N];
ll f[2][N][N],s[N],ans;
bool g[N][N];
int n;

ll GCD(ll a,ll b) {return !a?b:GCD(b%a,a);}

bool CMP(Node a,Node b) {return a.k<b.k;}

int main() {
    freopen("tree.in","r",stdin);
    freopen("tree.out","w",stdout);
    scanf("%d",&n);
    for (int i=1;i<=n;i++) scanf("%lld%lld",&a[i].k,&a[i].v);
    sort(a+1,a+n+1,CMP);
    for (int i=1;i<=n;i++) {
        s[i]=s[i-1]+a[i].v;
        for (int j=1;j<=n;j++) g[i][j]=GCD(a[i].k,a[j].k)>1ll?1:0;
    }
    for (int k=0;k<2;k++)
        for (int i=0;i<=n;i++)
            for (int j=0;j<=n;j++) f[k][i][j]=-Inf;
    for (int i=2;i<=n;i++) if (g[i][i-1]) f[1][i][i]=a[i].v;
    for (int i=1;i<n;i++) if (g[i][i+1]) f[0][i][i]=a[i].v;
    for (int len=1;len<n;len++)
        for (int l=1,r=l+len;r<=n;l++,r++)
            for (int j=l;j<=r;j++) {
                ll t=0;
                if (j==l) t=f[1][l+1][r];
                else
                if (j==r) t=f[0][l][r-1];
                else t=f[0][l][j-1]+f[1][j+1][r];
                if (l>1&&g[j][l-1]) f[1][l][r]=max(f[1][l][r],t+s[r]-s[l-1]);
                if (r<n&&g[j][r+1]) f[0][l][r]=max(f[0][l][r],t+s[r]-s[l-1]);
            }
    ans=f[1][2][n];
    for (int i=2;i<n;i++) ans=max(ans,f[0][1][i-1]+f[1][i+1][n]);
    ans=max(ans,f[0][1][n-1]);ans+=s[n];
    if (ans>=0) printf("%lld",ans);
    else printf("-1");
}