The meaning of problems: the distance defined between two points from one point to another point through the sum (including the two points), provided tuple (x, y) of two disjoint path, a length is x, a length y, asked tuple (x, y) there have been many times?
Ideas: jls direct on to explain:
Base diameter exercises.
Consider judge can not appear. On Jie: take any of a diameter of the tree, if it occurs, then there must be a scheme that are used at both ends of a diameter. Proved to be very simple: an endpoint is assumed that there is not used, consider two straight four endpoints, one endpoint will be able to move to the upper end of this diameter, since the diameter of the tree is a long path, so this movement must It does not reduce the length of the path.
Consider for each length, so that all values less than or equal to meet the requirements of long can also be satisfied. These values are calculated in consideration of the diameter: the first case, the diameter of the ends belong to different paths, then the endpoint can enumerate a diameter that path at a point diametrically, so that a large path length is the diameter of the end point to the point and this distance plus the length extending to the outer diameter of the large length of the other path may similarly request, wherein the large extension length can be calculated by one of DFS. The second case is that there is a path to this diameter, then the other path is the long path lengths beyond this diameter, this just put points on the diameter are deleted and then seek again diameter on the line.
The total time complexity is O (n).
Code:
#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int maxn = 100010;
vector<int> G[maxn];
int mp[maxn], mx_dis[maxn], f[maxn],Mx[maxn];
bool v[maxn], v1[maxn];
int root, mx, now;
vector<int> a;
void add(int x, int y) {
G[x].push_back(y);
G[y].push_back(x);
}
void dfs(int x, int fa, int dis, bool flag) {
f[x] = fa;
v1[x] = flag;
if(dis > mx) {
mx = dis;
now = x;
}
for (auto y : G[x]) {
if(y == fa || v[y]) continue;
dfs(y, x, dis + 1, flag);
}
}
int dfs1(int x, int fa) {
int mx = 0;
for (auto y : G[x]) {
if(v[y] || y == fa) continue;
mx = max(mx, dfs1(y, x));
}
return mx + 1;
}
int main() {
int T, x, y, n;
// freopen("6686in.txt", "r", stdin);
// freopen("out1.txt", "w", stdout);
scanf("%d", &T);
while(T--) {
scanf("%d", &n);
a.clear();
for (int i = 1; i <= n; i++)
G[i].clear();
for (int i = 1; i < n; i++) {
scanf("%d%d", &x, &y);
add(x, y);
}
for (int i = 1; i <= n; i++)
v[i] = v1[i] = mp[i] = f[i] = mx_dis[i] = 0;
root = 1, mx = 0;
dfs(root, 0, 1, 0);
root = now, mx = 0;
dfs(root, 0, 1, 0);
for (int i = now; i; i = f[i]) {
v[i] = 1;
a.push_back(i);
}
for (auto x : a) {
mx_dis[x] = dfs1(x, 0);
}
for (int i = 1; i <= n; i++) v1[i] = 0;
int tmp = 0;
for (int i = 1; i <= n; i++) {
if(v[i] == 0 && v1[i] == 0) {
now = mx = 0;
dfs(i, 0, 1, 0);
dfs(now, 0, 1, 1);
tmp = max(tmp, mx);
}
}
mp[a.size()] = tmp;
mp[tmp] = a.size();
Mx[a.size()] = 0;
for (int i = a.size() - 2; i >= 0; i--) {
Mx[i + 1] = max(Mx[i + 2], (int)a.size() - i - 2 + mx_dis[a[i + 1]]);
}
for (int i = 0; i < a.size() - 1; i++) {
int x = i + mx_dis[a[i]];
int y = Mx[i + 1];
mp[x] = max(mp[x], y);
mp[y] = max(mp[y], x);
}
LL ans = 0;
mx = 0;
for (int i = a.size(); i >= 1; i--) {
mx = max(mx, mp[i]);
ans += mx;
}
printf("%lld\n", ans);
}
}