P4408 [NOI2003] The diameter of the tree

Title

Portal P4408 [NOI2003] Children who play truant

answer

Set path $(C,A)$ and path$(A,$The first intersection of$B$$)$ is$D$ , the problem is equivalent to starting from$A$ starts and finds two paths$(A,B),(A,C)$ 且满足 $dist(A,D)\le dist(D,B)$，使 $dist(A,B)+dist(A,C)$ Maximum. Let$A,B$ is the end point of a certain path, then the answer is$$dist(A,B)+dist(C,D)+\min [dist(A,D),dist(D,B)],D\in path(A,B)$$ Let$(A,D)$ or$(D,B)$ is part of the bifurcation of the diameter of the tree. According to the longest diameter, replacing it with the diameter of the tree will not make the answer worse. Then$DFS$ solves the diameter of a certain tree, again$DFS$ solves the longest distance from each point on the diameter to its subtree node, and updates the answer by enumerating the points on the path. Total time complexity$O\left(N\right)$。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 200005;
int N, M, pre[maxn], num, rec[maxn];
int tot, head[maxn], to[maxn << 1], cost[maxn << 1], nxt[maxn << 1];
ll mx[maxn], ds[maxn];
bool in[maxn];

inline void add(int x, int y, int z)
{
    
    
    to[++tot] = y, cost[tot] = z, nxt[tot] = head[x], head[x] = tot;
}

void dfs(int x, int f, ll w, int &t)
{
    
    
    ds[x] = w, pre[x] = f;
    if (ds[x] >= ds[t])
        t = x;
    for (int i = head[x]; i; i = nxt[i])
    {
    
    
        int y = to[i], z = cost[i];
        if (y != f)
            dfs(y, x, w + z, t);
    }
}

void dfs2(int x, int f, ll w)
{
    
    
    mx[x] = w;
    for (int i = head[x]; i; i = nxt[i])
    {
    
    
        int y = to[i], z = cost[i];
        if (!in[y] && y != f)
            dfs2(y, x, w + z), mx[x] = max(mx[x], mx[y]);
    }
}

int main()
{
    
    
    scanf("%d%d", &N, &M);
    for (int i = 1, x, y, z; i <= M; ++i)
        scanf("%d%d%d", &x, &y, &z), add(x, y, z), add(y, x, z);
    int s = 1, t = 0;
    dfs(s, 0, 0, t);
    s = t, t = 0;
    dfs(s, 0, 0, t);
    for (int i = t; i; i = pre[i])
        rec[++num] = i, in[i] = 1;
    for (int i = 1; i <= num; ++i)
        dfs2(rec[i], 0, 0);
    ll dm = ds[t], res = 0;
    for (int i = 1, p; i <= num; ++i)
        p = rec[i], res = max(res, dm + mx[p] + min(ds[p], dm - ds[p]));
    printf("%lld\n", res);
    return 0;
}