Codeforces Round #430 (Div. 2) C. Ilya And The Tree dfs+set
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题意:给你一颗以1位根的有根数,每个节点都有一个权值,对于每个节点他的价值为从它到根节点所经过所有节点权值的最大公约数,而对于每个节点的计算,都可以将
某个节点的权值变为0,问你每个节点的价值的最大值。
做法:DFS+set 具体看代码。
//china no.1#pragma comment(linker, "/STACK:1024000000,1024000000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define fOR(n,x,i) for(int i=n;i>=x;i--)#define fOr(n,x,i) for(int i=n;i>x;i--)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double#define ll long longtypedef long long LL;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=2e5+10;const int maxx=1e6+100;const double EPS=1e-8;const double eps=1e-8;const int mod=1e9+7;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );//cerr << "run time is " << clock() << endl;set<int >st[maxn];int a[maxn];vector<int>G[maxn];bool vis[maxn];void dfs(int v,int u,int G_CD)//G_CD代表到当前节点前,即父亲节点的每个点都使用的GCD { vis[v]=1;//每个节点用一次 //for(auto it=st[u].begin();it!=st[u].end();it++) foreach(it,st[u]) { st[v].insert(__gcd(*it,a[v]));//把父亲节点的所有情况都加上选择这个节点时 } st[v].insert(G_CD);//即不使用这个节点 G_CD=__gcd(G_CD,a[v]);//使用这个节点 st[v].insert(G_CD); FOr(0,G[v].size(),i) { int x=G[v][i]; if(!vis[x]) dfs(x,v,G_CD); }}int main(){ int n; scan_d(n); FOR(1,n,i) scan_d(a[i]); int x,y; FOr(1,n,i) { scan_d(x); scan_d(y); G[x].push_back(y); G[y].push_back(x); } dfs(1,0,0); FOR(1,n,i) { printf("%d ",*st[i].rbegin()); } puts("");}
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