• 求树的重心 DFS


    树的重心

    何谓重心

    树的重心:找到一个点,其所有的子树中最大的子树节点数最少,那么这个点就是这棵树的重心,删去重心后,生成的多棵树尽可能平衡。

    树的重心可以通过简单的两次搜索求出,第一遍搜索求出每个结点的子结点数量son[u],第二遍搜索找出使max{son[u],n-son[u]-1}最小的结点。

    实际上这两步操作可以在一次遍历中解决。对结点u的每一个儿子v,递归的处理v,求出son[v],然后判断是否是结点数最多的子树,处理完所有子结点后,判断u是否为重心。

    以牛客的一道题:A病毒感染:https://www.nowcoder.com/acm/contest/214/A

    //#pragma GCC optimize(3)
    //#pragma comment(linker, "/STACK:102400000,102400000")  //c++
    // #pragma GCC diagnostic error "-std=c++11"
    // #pragma comment(linker, "/stack:200000000")
    // #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
    // #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3)
    
    #include <algorithm>
    #include  <iterator>
    #include  <iostream>
    #include   <cstring>
    #include   <cstdlib>
    #include   <iomanip>
    #include    <bitset>
    #include    <cctype>
    #include    <cstdio>
    #include    <string>
    #include    <vector>
    #include     <stack>
    #include     <cmath>
    #include     <queue>
    #include      <list>
    #include       <map>
    #include       <set>
    #include   <cassert>
    
    using namespace std;
    #define lson (l , mid , rt << 1)
    #define rson (mid + 1 , r , rt << 1 | 1)
    #define debug(x) cerr << #x << " = " << x << "
    ";
    #define pb push_back
    #define pq priority_queue
    
    
    
    typedef long long ll;
    typedef unsigned long long ull;
    //typedef __int128 bll;
    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 '
    '
    
    #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)
    #define max3(a,b,c) max(max(a,b), c);
    //priority_queue<int ,vector<int>, greater<int> >que;
    
    const ll mos = 0x7FFFFFFF;  //2147483647
    const ll nmos = 0x80000000;  //-2147483648
    const int inf = 0x3f3f3f3f;
    const ll inff = 0x3f3f3f3f3f3f3f3f; //18
    const int mod = 1e9+7;
    const double esp = 1e-8;
    const double PI=acos(-1.0);
    const double PHI=0.61803399;    //黄金分割点
    const double tPHI=0.38196601;
    
    
    template<typename T>
    inline T read(T&x){
        x=0;int f=0;char ch=getchar();
        while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
        while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
        return x=f?-x:x;
    }
    
    
    /*-----------------------showtime----------------------*/
            int minn = inf;
            int n,m;
            const int maxn = 50009;
            vector<int>mp[maxn];
            int dp[maxn],d[maxn];
            void dfs(int u,int fa){
                dp[u] = 1;
                d[u] = 0;
                for(int i=0; i<mp[u].size(); i++){
                    int v = mp[u][i];
                    if(v == fa)continue;
                    dfs(v, u);
                    dp[u] += dp[v];
                    d[u] = max(d[u], dp[v]);
                }
    
                d[u] = max(d[u], n - dp[u]);
                minn = min(minn, d[u]);
            }
    int main(){
            scanf("%d%d", &n, &m);
            for(int i=1; i<=m; i++){
                int u,v;
                scanf("%d%d", &u, &v);
                mp[u].pb(v);    mp[v].pb(u);
            }
            dfs(1, -1);
           // debug(minn);
            for(int i=1; i<=n; i++){
                if(d[i] == minn){printf("%d ", i);}
            }
            printf("
    ");
            return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/ckxkexing/p/9825277.html
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