求树的重心 POJ 1655、POJ 3107 树形DP

树的重心：在树中删去重心后，剩下的子树的最大子树大小最小，则这个节点为重心

POJ 1655：直接求树的重心

思路：递归每个节点的时候，记录一下这个节点所有子树的大小和，然后记录最大子树的大小，维护最小的最大子树大小和相应的节点即可

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#define sf scanf
#define pf printf

using namespace std;
const int maxn = 20000 + 5,INF = 1e9;
int son[maxn],ans,ans_size;
struct Edge{
int v,pre;
}Edges[maxn * 2];
int head[maxn],tot,n;
void Init(){
memset(head,-1,sizeof head),tot = 0;
}
void Add(int u,int v){
Edges[tot].v = v;
Edges[tot].pre = head[u];
head[u] = tot++;
}
void DFS(int u,int fa){
int tmp = 0;
son[u] = 0;
for(int i = head[u];~i;i = Edges[i].pre){
int v = Edges[i].v;
if(v == fa) continue;
DFS(v,u);
tmp = max(tmp,son[v] + 1);
son[u] += son[v] + 1;
}
tmp = max(tmp,n - 1 - son[u]);
if(tmp < ans_size || (tmp == ans_size && u < ans)){
ans_size = tmp;
ans = u;
}
}
int main(){
int T;sf("%d",&T);
while(T--){
Init();
sf("%d",&n);
for(int i = 0;i < n - 1;++i){
int u,v;sf("%d %d",&u,&v);
Add(u,v);
Add(v,u);
}
memset(son,0,sizeof son);
ans_size = INF;
DFS(1,0);
pf("%d %d\n",ans,ans_size);
}
}

POJ 3107 输出所有的重心

思路：同上，不过要记录一下所有的节点

#include <cstring>
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#define sf scanf
#define pf printf
using namespace std;
const int maxn = 50000 + 5,INF = 2e9;
struct Edge{
int v,pre;
}Edges[maxn * 2];
int head[maxn],ans_size,tot,n,son[maxn];
vector<int> ans;
void init(){
memset(head,-1,sizeof head),tot = 0;
memset(son,0,sizeof son);
ans_size = INF;
}
void Add(int u,int v){
Edges[tot].v = v;
Edges[tot].pre = head[u];
head[u] = tot++;
}
void DFS(int u,int fa){
int tmp = 0;son[u] = 0;
for(int i = head[u];~i;i = Edges[i].pre){
int v = Edges[i].v;
if(v == fa) continue;
DFS(v,u);
son[u] += son[v] + 1;
tmp = max(tmp , son[v] + 1);
}tmp = max(tmp,n - 1 - son[u]);
if(tmp < ans_size){
ans.clear();ans.push_back(u);ans_size = tmp;
}else if(tmp == ans_size) ans.push_back(u);
}
int main(){
while(~sf("%d",&n)){
init();
for(int i = 1;i < n;++i){
int u,v;sf("%d %d",&u,&v);
Add(u,v);
Add(v,u);
}
DFS(1,0);
sort(ans.begin(),ans.end());
int len = ans.size();
for(int i = 0;i < len;++i) pf("%d%c",ans[i]," \n"[i + 1 == len]);
}
}