Codeforces Round #505 -D-Recovering BST（區間DP）
D. Recovering BST
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Dima the hamster enjoys nibbling different things: cages, sticks, bad problemsetters and even trees!
Recently he found a binary search tree and instinctively nibbled all of its edges, hence messing up the vertices. Dima knows that if Andrew, who has been thoroughly assembling the tree for a long time, comes home and sees his creation demolished, he'll get extremely upset.
To not let that happen, Dima has to recover the binary search tree. Luckily, he noticed that any two vertices connected by a direct edge had their greatest common divisor value exceed 11.
Help Dima construct such a binary search tree or determine that it's impossible. The definition and properties of a binary search tree can be found
Input
The first line contains the number of vertices nn (2≤n≤7002≤n≤700).
The second line features nn distinct integers aiai (2≤ai≤1092≤ai≤109) — the values of vertices in ascending order.
Output
If it is possible to reassemble the binary search tree, such that the greatest common divisor of any two vertices connected by the edge is greater than 11, print "Yes" (quotes for clarity).
Otherwise, print "No" (quotes for clarity).
Examples
input
Copy
6 3 6 9 18 36 108
output
Copy
Yes
input
Copy
2 7 17
output
Copy
No
input
Copy
9 4 8 10 12 15 18 33 44 81
output
Copy
Yes
Note
The picture below illustrates one of the possible trees for the first example.
The picture below illustrates one of the possible trees for the third example.
題意：給你n個節點，每個節點有一個權值，兩個點可以連邊當且僅當這兩個點的gcd>1,問你這n個點能否構成一個二叉搜尋樹（每個節點最多有兩個兒子，且左兒子小於右兒子），輸入為遞增順序。
題解：我們考慮區間dp，設兩個陣列L[l][r]和R[l][r]分別表示[l,r]這一區間內能否構成左子樹和能否構成右子樹，然後簡單轉移即可。（詳見程式碼）
#include<stdio.h>
#define maxn 710
int n,a[maxn],L[maxn][maxn];
int R[maxn][maxn],g[maxn][maxn];
int gcd(int a,int b)
{
if(b==0)
return a;
return gcd(b,a%b);
}
int main(void)
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]),L[i][i]=R[i][i]=1;
for (int i=1;i<=n;++i)
for (int j=1;j<=n;++j)
g[i][j]=gcd(a[i],a[j]);
for (int l=n;l>=1;l--)
for (int r=l;r<=n;r++)
for (int k=l;k<=r;k++)
if (L[l][k] && R[k][r])
{
if (l==1 && r==n)
{
printf("Yes\n");
return 0;
}
if (g[l-1][k]>1) R[l-1][r]=1;
if (g[k][r+1]>1) L[l][r+1]=1;
}
printf("No\n");
return 0;
}