树的直径

U81904 【模板】树的直径

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有以下两种办法求得树的直径：

1. 两遍 DFS（或BFS）：可以记录点
2. 树形 DP：允许有负边权

首先是 两遍 DFS 可求路径

int head[N], cnt;
struct Edge {
    int from, to, nxt;
    ll val;
}e[N << 1];
void add(int u, int v, ll val){
    e[++cnt].from = u;
    e[cnt].to = v;
    e[cnt].nxt = head[u];
    head[u] = cnt;
    e[cnt].val = val;
}

ll dis[N], fa[N];
void dfs(int u, int father, ll d){
    fa[u] = father;
    dis[u] = d;
    for(int i = head[u]; i != 0; i = e[i].nxt){
        int v = e[i].to;
        if(v == father) continue;
        dfs(v, u, d + e[i].val);
    }

}
ll max_len(int &beg, int &ed){
    dfs(1, -1, 0);
    beg = 1;
    for(int i = 1; i <= n; i++){
        if(dis[i] > dis[beg]) beg = i;
    }
    dfs(beg, -1, 0);
    for(int i = 1; i <= n; i++){
        if(dis[i] > dis[ed]) ed = i;
    }
    return dis[ed];
}

2024.7.7 补充：主函数调用时需要初始化 beg 和 ed，如下 , abc361 就 re 了一发

int x = 0, y = 0;
ll res = max_len(x, y);

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然后是 树形DP 做法，可以维护负边权

点击查看代码

int head[N], cnt;
struct Edge {
    int from, to, nxt;
    ll val;
}e[N << 1];
void add(int u, int v, ll val){
    e[++cnt].from = u;
    e[cnt].to = v;
    e[cnt].nxt = head[u];
    head[u] = cnt;
    e[cnt].val = val;
}
ll dp[N], maxlen;
bool vis[N];

void dfs(int u){
    vis[u] = true;
    for(int i = head[u]; i != 0; i = e[i].nxt){
        int v = e[i].to;
        if(vis[v]) continue;
        dfs(v);
        maxlen = max(maxlen, dp[u] + dp[v] + e[i].val);
        dp[u] = max(dp[u], dp[v] + e[i].val);
    }
    return ;
}