题意:
有一个递增序列,问能不能构建出一颗每条边的端点值都不互质的二叉排序树。
思路:
区间DP,但是和常见的区间DP不一样,
这里dp【i】【j】表示的是区间【i,j】能否以i为根建立一个小二叉排序树。
所以可以得到dp【i】【j】 为true, 要求在【i+1,j】中有一个k,dp【k】【i+1】和dp【k】【j】都为true。
或者在i点的左边取件中,即要求在【j】【i-1】中有一个k,dp【k】【j】和dp【k】【i-1】都为true。
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <cmath> #include <queue> #include <list> #include <map> #include <set> using namespace std; //#pragma GCC optimize(3) //#pragma comment(linker, "/STACK:102400000,102400000") //c++ #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl ‘\n‘ #define OKC ios::sync_with_stdio(false);cin.tie(0) #define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行 #define REP(i , j , k) for(int i = j ; i < k ; ++i) //priority_queue<int ,vector<int>, greater<int> >que; const ll mos = 0x7FFFFFFFLL; //2147483647 const ll nmos = 0x80000000LL; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3fLL; //18 const int mod = 998244353; const double PI=acos(-1.0); // #define _DEBUG; //*// #ifdef _DEBUG freopen("input", "r", stdin); // freopen("output.txt", "w", stdout); #endif /*-----------------------showtime----------------------*/ const int maxn = 800; ll a[maxn],mp[maxn][maxn],dp[maxn][maxn]; ll gcd(ll a,ll b){ if(b==0)return a; return gcd(b,a%b); } int main(){ int n; scanf("%d", &n); for(int i=1; i<=n; i++){ scanf("%I64d", &a[i]); } for(int i=1; i<=n; i++){ for(int j=1; j<=n; j++) { if(i==j)dp[i][j] = 1; mp[i][j] = (gcd(a[i],a[j]) == 1?0:1); } } for(int len = 1; len <= n; len++){ for(int i=1; i<=n; i++){ int le = i - len; if(le >= 1){ for(int k = le ; k < i; k++){ if(dp[k][le] && dp[k][i-1] && mp[k][i]){ dp[i][le] = 1; break; } } } int ri = i + len; if(ri <= n){ for(int k = i+1; k <= ri ; k++){ if(dp[k][i+1] && dp[k][ri] && mp[i][k]){ dp[i][ri] = 1; break; } } } } } for(int i=1; i<=n; i++){ if(dp[i][1] && dp[i][n]){ puts("Yes"); return 0; } } puts("No"); return 0; }
CF1025D
原文地址:https://www.cnblogs.com/ckxkexing/p/9531045.html
时间: 2024-10-12 12:12:50