『算法-ACM 竞赛-图论』树的直径-DFS+树形 DP 模板

图论–树的直径–DFS+树形 DP 模板

#include <iostream>
#include <cstring>
using namespace std;

//maxv：源点能到的最远点，maxdis:最远点对应的距离,
const int maxn = 1e4 + 5;
struct Edge { int to, next, w; }edges[2 * maxn];
int head[maxn], maxdis,maxv, tot;

void add(int u, int v, int w) {
    edges[tot] = { v, head[u], w };
    head[u] =tot++;
}

void dfs(int u, int f, int Val) {
    if (maxdis < Val){
        maxdis = Val;
        maxv = u;
    }
    for (int e = head[u]; e != -1; e = edges[e].next) {
        int v = edges[e].to, w = edges[e].w;
        if (v == f) continue;  //父节点已经访问过，防止重复遍历，相反孩子不会重复遍历。
        dfs(v, u, Val + w);
    }
}

int main()
{
    int e, u, v, w, s;
    cin >> e;
    memset(head, -1, sizeof(head));
    for (int i = 1; i <= e; i++) {
        cin >> u >> v >> w;
        add(u, v, w), add(v, u, w);
    }
    dfs(1, -1, 0); //从结点1开始遍历，找到最远点maxv及对应的最远距离maxdis
    maxdis = 0;
    cout <<maxv<<endl;//输出直径的第一个端点
    dfs(maxv, -1, 0);//从结点maxv开始遍历，找到最远点对应的距离maxdis
    cout << maxdis << endl; //輸出树的直径
    cout <<maxv<<endl;//输出树的直径的第二个端点
    return 0;
}


#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const MAX = 100005;
int head[MAX], dp[MAX][2];
int n, s, cnt, ans;
struct EDGE
{
  int v, w, next;
} e[MAX];
void Add(int u, int v, int w)
{
  e[cnt].v = v;
  e[cnt].w = w;
  e[cnt].next = head[u];
  head[u] = cnt++;
}
void DFS(int u, int fa)
{
  dp[u][0] = dp[u][1] = 0;
  for (int i = head[u]; i != -1; i = e[i].next)
  {
    int v = e[i].v;
    int w = e[i].w;
    if (v != fa)
    {
      DFS(v, u);
      if (dp[u][0] < dp[v][0] + w)
      {
        int tmp = dp[u][0];
        dp[u][0] = dp[v][0] + w;
        dp[u][1] = tmp;
      }
      else if (dp[u][1] < dp[v][0] + w)
        dp[u][1] = dp[v][0] + w;
    }
  }
  ans = max(ans, dp[u][1] + dp[u][0]);
  return;
}
int main()
{
  cnt = 0;
  ans = 0;
  memset(head, -1, sizeof(head));
  scanf("%d %d", &n, &s);
  int sum = 0;
  for (int i = 0; i < n - 1; i++)
  {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    Add(u, v, w);
    Add(v, u, w);
    sum += 2 * w;
  }
  DFS(s, -1);
  printf("%d\n", ans);
}