Multi-field correction ninth HDU 1007 Rikka with Travels - DP + tree root change

Topic links: point I ah ╭ (╯ ^ ╰) ╮

Subject to the effect:

    Unrooted tree defined $L(a,b)$ is from tree $a$ to $b$ route points
    calculated $pairs (l_1, l_2)$ ， $L(a, b)=l_1, L(c,d)=l_2$
    And from $a$ to $b$ of path $c$ to $d$ path disjoint

Problem-solving ideas:

    Compute $pairs (l_1, l_2)$ , If found to meet $pair(x, y)$
    is also satisfied $pairs(i, y)，i \in (1,x)$
    simply requires that for each $x$ , the largest $Y$ can
    then $mx[i] = max(mx[i], mx[i+1])$ , may be $O (n)$ calculated answer

    Consider the path does not intersect, the edges can be enumerated, with this edge to split the tree
    and calculating the diameter of two subtrees $x$ and $Y$ , corresponding to the above
    diameter can be calculated directly subtree tree $DP$， $dp[rt][0]$ represents $rt$ maximum depth
     $dp[rt][1]$ represents $rt$ maximum diameter is selected within a maximum of two subtrees $dp[son][0]$ update to

---

    Then consider how to enumerate the edge, thinking to change the root $dp$ in from father $rt$ change to $son$ time
    first delete $rt$ to $son$ side update $rt$ information at this time to achieve the purpose of enumeration side
    is calculated directly in the process of changing the root of the answer can
    change root similar questions:HDU multi-school eighth-field 1006 Acesrc and Travel - tree DP + change root

Core: DP + tree root transducer (enumeration side)

#include<bits/stdc++.h>
#define rint register int
#define deb(x) cerr<<#x<<" = "<<(x)<<'\n';
#define x first
#define y second
using namespace std;
typedef long long ll;
using pii = pair <int,int>;
const int maxn = 1e5 + 5;
int T, n, dp[maxn][2], mx[maxn];
vector <int> g[maxn];
map <pii, int> m[maxn][2];
map <pii, int>::iterator it;

inline void add0(int u, int v){
    m[u][0][{dp[v][0], v}]++;
    while(m[u][0].size()>3) m[u][0].erase(m[u][0].begin());
}
inline void add1(int u, int v){
    m[u][1][{dp[v][1], v}]++;
    while(m[u][1].size()>3) m[u][1].erase(m[u][1].begin());
}

inline void get(int rt){
    dp[rt][0] = dp[rt][1] = 1;
    if(m[rt][0].size()){
        it = m[rt][0].end(), --it;
        dp[rt][0] = dp[rt][1] = it->x.x + 1;
        if(m[rt][0].size() > 1){
            int tmp = it->x.x;
            --it;
            dp[rt][1] = max(dp[rt][1], tmp + it->x.x + 1);
        }
    }
    if(m[rt][1].size()){
        it = m[rt][1].end(), --it;
        dp[rt][1] = max(dp[rt][1], it->x.x);
    }
}

inline void dfs(int u, int fa){
    m[u][0].clear(), m[u][1].clear();
    for(auto v : g[u]){
        if(v == fa) continue;
        dfs(v, u);
        add0(u, v);
        add1(u, v);
    }
    get(u);
}

inline void gao(int x, int y){
    mx[x] = max(mx[x], y);
    mx[y] = max(mx[y], x);
}

inline void move(int rt, int son){
    m[rt][0].erase( {dp[son][0], son} );
    m[rt][1].erase( {dp[son][1], son} );
    get(rt);
    gao(dp[rt][1], dp[son][1]);
    add0(son, rt);
    add1(son, rt);
    get(son);
}

inline void dfs1(int rt, int fa){
    for(auto son : g[rt]){
        if(son == fa) continue;
        move(rt, son);
        dfs1(son, rt);
        move(son, rt);
    }
}

int main() {
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        memset(mx, 0, sizeof(mx));
        for(int i=1; i<=n; i++) g[i].clear();
        for(int i=1, u, v; i<n; i++){
            scanf("%d%d", &u, &v);
            g[u].push_back(v);
            g[v].push_back(u);
        }
        dfs(1, 0);
        dfs1(1, 0);
        ll ans = 0;
        for(int i=n; i; i--) mx[i] = max(mx[i], mx[i+1]), ans += mx[i];
        printf("%lld\n", ans);
    }
}