POJ 2378 Tree Cutting (DFS)

时间:2023-03-08 19:37:49

题目链接:http://poj.org/problem?id=2378

一棵树,去掉一个点剩下的每棵子树节点数不超过n/2。问有哪些这样的点,并按照顺序输出。

dfs回溯即可。

 //#pragma comment(linker, "/STACK:102400000, 102400000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
typedef pair <int, int> P;
const int N = 1e4 + ;
vector <int> ans;
struct Edge {
int next, to;
}edge[N << ];
int d[N], head[N], cnt, n; inline void add(int u, int v) {
edge[cnt].next = head[u];
edge[cnt].to = v;
head[u] = cnt++;
} void dfs(int u, int p) {
bool ok = true;
d[u] = ;
for(int i = head[u]; ~i; i = edge[i].next) {
int v = edge[i].to;
if(v == p)
continue;
dfs(v, u);
if(d[v] * > n)
ok = false;
d[u] += d[v];
}
if((n - d[u]) * > n)
ok = false;
if(ok)
ans.push_back(u);
} int main()
{
while(~scanf("%d", &n)) {
int u, v;
memset(head, -, sizeof(head));
cnt = ;
ans.clear();
for(int i = ; i < n; ++i) {
scanf("%d %d", &u, &v);
add(u, v);
add(v, u);
}
dfs(, -);
sort(ans.begin(), ans.end());
for(int i = ; i < ans.size(); ++i) {
printf("%d\n", ans[i]);
}
}
return ;
}