树的直径【树的重心 / 树的质心】（填坑中）

定义：

对于一棵树n个节点的无根树，找到一个点，将树变成以该点为根的有根树，而重心则是删除某节点 Node 之后能使得最大子树的结点数最小的节点。

性质：

　　1.删除重心后所得的所有子树，节点数不超过原树的1/2，一棵树最多有两个相邻的重心；

　　2.树中所有节点到重心的距离之和最小，如果有两个重心，那么他们距离之和相等；

　　3.两个树通过一条边合并，新的重心在原树两个重心的路径上；

　　4.树删除或添加一个叶子节点，重心最多只移动一条边；

1）Balancing Act

题意：

对于N个节点N-1条边相连接的树，问这棵树的重心和对应的最大子树的节点数为多少

题解：

#include <iostream>#include <map>#include <vector>#include <queue>#include <string>#include <t>#include <algorithm>#include <cstring>#include <string>#include <vector>#include <map>#include <t>#include <list>#include <deque>#include <queue>#include <stack>#include <cstdlib>#include <cstdio>#include <cmath>#include <iomanip>#pragma GCC optimize(2)using namespace std;const int inf=0x3f3f3f3f;#define ll long long#define ull unsigned long longconst int mod = 1e9+7;/// 998244353;const int mxn = 2e4 +7;const int N = 8e4 + 7 ;int _ , n , m , t , k , ans , cnt , si ;template <class T>void rd(T &x){    T flag = 1 ; x = 0 ; char ch = getchar() ;    while(!isdigit(ch)) { if(ch=='-') flag = -1; ch = getchar(); }    while(isdigit(ch)) { x = (x<<1) + (x<<3) + (ch^48); ch = getchar(); }    x*=flag;}vector<int> G[mxn] ;int deg[mxn] , mn_son = 0 , mn_node = 0 , head[mxn<<1] , to[mxn<<1] , nx[mxn<<1] ; void add(int u,int v) { to[cnt] = v , nx[cnt] = head[u] , head[u] = cnt++; }void DFS(int x,int par){    deg[x]++ ;    int mx_son = 0 ;    for(int i=head[x];~i;i=nx[i]){        int v = to[i] ;        if(v==par) continue ;        DFS(v,x);        deg[x] += deg[v] ;        mx_son = max(mx_son , deg[v]);    }    mx_son = max(mx_son , n-deg[x]) ;    if(mn_son > mx_son)        mn_node = x , mn_son = mx_son ; }void solve(){    for(rd(t);t;t--){        rd(n) ;         mn_node = 0 , mn_son = mod ; cnt = 0 ;        memt(head,-1,sizeof(head));        memt(deg,0,sizeof(deg));        for(int i=1,u,v;i<n;i++){            rd(u) , rd(背诵法v) ;            add(u,v) , add(v,u) ;        }        DFS(1,0);        printf("%d %d\n",mn_node,mn_son );    }}int main(){    ios::sync_with_stdio(fal); cin.tie(0) ; cout.tie(0);    solve();}

2）Cotree

题意：

由N个点构成两棵树，问在两颗树之间连接一条边之后，各点之间距离和的最小值为多少

题解：

进行两次DFS找到两颗树的重心能源动力专业，将两个重心连接起来，再进行一次DFS求出距离和即可

#include <iostream>#include <map>#include <vector>#include <queue>#include <string>#include <t>#include <algorithm>#include <cstring>#include <string>#include <vector>#include <map>#include <t>#include <list>#include <deque>#include <queue>#include <stack>#include <cstdlib>#include <cstdio>#include <cmath>#include <iomanip>#pragma GCC optimize(2)#define ll long long#define ull unsigned long longusing namespace std;template <class T>void rd(T &x){    x = 0 ;T flag = 1光反应 ;char ch = getchar();    while(!isdigit(ch)) { if(ch=='-') flag = -1 ; ch = getchar() ; }    while(isdigit(ch)) { x = (x<<1) + (x<<3) + (ch^48) ; ch = getchar(); }    x*= flag ;}const int INF=0x3f3f3f3f;const int mxn = 1e5+7;ll t,n,m,k,cnt,node_1,node_2,ans_1,ans_2,mn_son = 0 , mn_node = 0 , res , ans ;int head[mxn<<2] , to[mxn<<2] , nx[mxn<<2] ,deg[mxn<<2] ;int vis[mxn<<2] ;void add(int u,int v){ to[cnt] = v , nx[cnt] = head[u] , head[u] = cnt++; }void DFS_INIT(int x) /// 计算两颗子树的节点数量{    if(vis[x]) return ;    res++; vis[x] = 1 ;    for(int i=head[x];~i;i=nx[i]) if(!vis[to[i]]) DFS_INIT(to[i]); /// }void DFS(int x,int par,int n){    deg[x]=1; int mx_son = 0 ;    for(int i=head[x];~i;i=nx[i]){        int v = to[i] ;        if(v==par) continue;        DFS(v,x,n);        deg[x] += deg[v] ;        mx_son = max(mx_son , deg[v]);/// 子树的最大节点个数    }    mx_son = max(mx_son,n-deg[x]); /// 子树的最大节点个数    if( mx_son < mn_son ){        mn_son = mx_son , mn_node = x ;    }}void DFS_END(int x,int par){    deg[x] = 1 ;    for(int i=head[x];~i;i=nx[i]){        int v = to[i] ;        if(v==par) continue;        DFS_END(v,x) ;        deg[x] += deg[v];        ans+=(ll)(deg[v])*(ll)(n-deg[v]);    }}int main(){    while(~scanf("%lld",&n)){        cnt = 0 ;        memt(deg,0,sizeof(deg));        memt(head,-1,sizeof(head));        for(int i=1,u,v;i<=n-2;i++){            rd(u) , rd(v) ;            add(u,v) , add(v,u) ;        }        memt(vis,0,sizeof(vis));        int n1 = 0 , n2 = 0 ; res = 0 ;        DFS_INIT(1) , node_1 = 1 , n1 = res , res = 0 ;        for(int i=1;i<=n;i++){            if(!vis[i]) {                DFS_INIT(i) , n2 = res , node_2 = i , res = 0 ;                break;            }        }        mn_son = INF ; int e1 , e2 ;        DFS(node_1,0,n1) , e1 = mn_node ,mn_son = INF ;        DFS(node_2,0,n2) , e2 = mn_node ,mn_son = INF ;        ///cout<<e1<<" -- "<<e2<<endl;        add(e1,e2) , add(e2,e1);        ans = 0 ; DFS_END(1,0);        printf("%lld\n",ans);    }    return 0;}

3）C. Link Cut Centroids

题意：

在N个点的树中，通过删除一条边，连接一条边使得新树只有一个重心，并输出删除和连接的点

题解：

通过DFS判断存在几个重心，然后找到一个重心的子树的叶子节点，如果是一个重心，那么输出两遍叶子节点和连接的点；如果是两个重心，则将找到的叶子节点连接到另一个重心上

#include <iostream>#include <map>#include <vector>#include <queue>#include <string>#include <t>#include <alg隋唐英雄罗成orithm>#include <cstring>#include <string>#include <vector>#include <map>#include <t>#include <list>#include <deque>#include <queue>#include <stack>#include <cstdlib>#include <cstdio>#include <cmath>#include <iomanip>#pragma GCC optimize(2)using namespace std;const int inf=0x3f3f3f3f;#define ll long long#define ull unsigned long longconst int mod = 1e9+7;/// 998244353;const int 项目进度管理mxn = 1e5 +7;int _ , n , m , t , k , ans , cnt , si , res ;template <class T>void rd(T &x){    T flag = 1 ; x = 0 ; char ch = getchar() ;    while(!isdigit(ch)) { if(ch=='-') flag = -1; ch = getchar(); }    while(isdigit(ch)) { x = (x<<1) + (x<<3) + (ch^48); ch = getchar(); }    x*=flag;}vector<int> G[mxn] ;int deg[mxn] , _son = 0 , _node = 0 , head[mxn<<1] , to[mxn<<1] , nx[mxn<<1] , id[5] , node ; void add(int u,int v){ to[cnt] = v , nx[cnt] = head[u] , head[u] = cnt++ ; }void DFS(int u,int par){    deg[u] = 1 ; int son = 0 ;    for(int i=head[u];~i;i=nx[i]){        int v = to[i] ;        if(v==par) continue ;        DFS(v,u);        deg[u] += deg[v] ;        son = max(son,deg[v]);    }    son = max(son,n-deg[u]);    if( son < _son )        res = 0 , _son = son , _node = u , id[res++] = u ;    el if(son==_son){        id[res++] = u ;    }}void _DFS(int x,int par){    if(deg[x]==1) {        node = x ;        return ;    }    for(int i=head[x];~i;i=nx[i]){        int v = to[i] ;        if(v==par || v==id[1]) continue ;        _DFS(v,x);    }}void solve(){    for(rd(t);t;t--){        rd(n);        memt(head,-1,sizeof(head));        memt(deg,0,sizeof(deg));        for(int i=1,u,v;i<n;i++){            rd(u) , rd(v) ;            add(u,v) , add(v,u) ;        }        _son = inf ;        DFS(1,0);        if(res==1){            for(int j=1;j<=n;j++){                if(deg[j]==1){                    for(int i=head[j];~i;i=nx[i]){                        printf("%d %d\n%d %d\n",j,to[i],j,to[i]);                        break;                    }                    break;                }            }        } el {            _DFS(id[0],0);            for(int i=head[node];~i;i=nx[i]){                printf("%d %d\n%d %d\n",node,to[i],id[1],node);                break;            }        }    }}int main(){    /// freopen("input.in","r",stdin) ; freopen("output.in","w",stdout) ;    ios::sync_with_stdio(fal); cin.tie(0) ; cout.tie(0);    solve();}

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