传送门
Luogu
解题思路
显然的贪心策略,因为每次都要尽量使得删点后的收益最大。
我们可以求出树的直径(因为树上的任意一个节点与其距离最远的点一定是直径的端点)。
然后我们对于所有不是直径上的点,从叶子开始,从下往上删点,最后再由深而浅删掉直径。
最后输出答案即可。
细节注意事项
- 有些地方的计算不要写错式子之类的
参考代码
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <ctime>
#include <queue>
#define rg register
using namespace std;
template < typename T > inline void read(T& s) {
s = 0; int f = 0; char c = getchar();
while (!isdigit(c)) f |= (c == '-'), c = getchar();
while (isdigit(c)) s = s * 10 + (c ^ 48), c = getchar();
s = f ? -s : s;
}
typedef long long LL;
const int _ = 200010;
const int __ = 400010;
int tot, head[_], nxt[__], ver[__];
inline void Add_edge(int u, int v)
{ nxt[++tot] = head[u], head[u] = tot, ver[tot] = v; }
int n, rt, lf, mark[_], fa[_], dep[_];
queue < pair < int, int > > Q;
inline void dfs1(int u, int f) {
for (rg int i = head[u]; i; i = nxt[i]) {
int v = ver[i]; if (v == f) continue;
dep[v] = dep[u] + 1, dfs1(v, u);
}
}
inline void dfs2(int u, int f) {
for (rg int i = head[u]; i; i = nxt[i]) {
int v = ver[i]; if (v == f) continue;
fa[v] = u, dfs2(v, u), mark[u] |= mark[v];
}
}
LL ans = 0;
inline void dfs3(int u, int f, int lca) {
for (rg int i = head[u]; i; i = nxt[i]) {
int v = ver[i]; if (v == f) continue;
dfs3(v, u, mark[v] ? v : lca);
}
if (!mark[u]) {
if (dep[u] > dep[u] + dep[lf] - 2 * dep[lca])
ans += dep[u], Q.push(make_pair(rt, u));
else
ans += dep[u] + dep[lf] - 2 * dep[lca], Q.push(make_pair(lf, u));
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("in.in", "r", stdin);
#endif
read(n);
for (rg int u, v, i = 1; i < n; ++i)
read(u), read(v), Add_edge(u, v), Add_edge(v, u);
dep[1] = 0, dfs1(1, 0);
rt = 1;
for (rg int i = 1; i <= n; ++i)
if (dep[i] > dep[rt]) rt = i;
dep[rt] = 0, dfs1(rt, 0);
lf = rt;
for (rg int i = 1; i <= n; ++i)
if (dep[i] > dep[lf]) lf = i;
mark[lf] = 1;
dfs2(rt, 0);
dfs3(rt, 0, rt);
for (rg int i = lf; i != rt; i = fa[i])
ans += dep[i], Q.push(make_pair(rt, i));
printf("%lld\n", ans);
while (!Q.empty()) {
pair < int, int > x = Q.front(); Q.pop();
printf("%d %d %d\n", x.first, x.second, x.second);
}
return 0;
}
完结撒花 \(qwq\)
原文地址:https://www.cnblogs.com/zsbzsb/p/11745871.html
时间: 2024-10-30 00:50:46