一、视频讲解

视频讲解

二、暴力代码

//暴力代码：DFS
#include<bits/stdc++.h>
#define endl '\n'
#define deb(x) cout << #x << " = " << x << '\n';
#define INF 0x3f3f3f3f
#define int long long
using namespace std;
const int N = 2e5 + 10;typedef pair<int,int> pii;map<pii, int>st;//记录从{x, y}的距离是多少
int a[N];vector<pii>edge[N];//存图//s表示你要求的路径的起点
//v表示你要求的路径的终点
//u表示你当前走到了哪个点
//father表示你当前这个点的父亲节点是谁。避免重复走造成死循环
//sum表示从s走到u的路径花费总和。
bool dfs(int s, int u, int father, int v, int sum)
{if(u == v){st[{s, v}] = sum;st[{v, s}] = sum;// cout << s << " " << v << " " << sum << endl;return true;}for(int i = 0; i < edge[u].size(); i ++){int son = edge[u][i].first;if(son == father)continue;int w = edge[u][i].second;if(dfs(s, son, u, v, sum + w))return true;}return false;
}void solve()
{int n, k;cin >> n >> k;for(int i = 0; i < n - 1; i ++){int x, y, t;cin >> x >> y >> t;edge[x].push_back({y, t});edge[y].push_back({x, t});}for(int i = 0; i < k; i ++)cin >> a[i];//求出完整路线的总花费//O(k * n)int ans = 0;for(int i = 0; i < k - 1; i ++){dfs(a[i], a[i], -1, a[i + 1], 0);ans += st[{a[i] ,a[i + 1]}];}for(int i = 0; i < k; i ++){int tmp = ans;if(i == 0)tmp -= st[{a[i], a[i + 1]}];else if(i == k - 1)tmp -= st[{a[i - 1], a[i]}];else{tmp -= st[{a[i - 1], a[i]}];tmp -= st[{a[i], a[i + 1]}];dfs(a[i - 1], a[i - 1], -1, a[i + 1], 0);tmp += st[{a[i - 1], a[i + 1]}];}cout << tmp << endl;}}signed main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int t = 1;//cin >> t;while(t--)solve();
}

三、正解代码

//景区导游：树上前缀和 + 最近公共祖先
#include<bits/stdc++.h>
#define int long long
using namespace std;
typedef pair<int, int> pii;
const int N = 1e5 + 10;
int a[N], siz[N], dep[N], fa[N], son[N], top[N];
int sum[N];
int n, k;
vector<pii>edge[N];void dfs1(int u, int father)
{siz[u] = 1, dep[u] = dep[father] + 1;fa[u] = father;for(int i = 0; i < edge[u].size(); i ++){int s = edge[u][i].first;if(s == father)continue;dfs1(s, u);siz[u] += siz[s];if(siz[son[u]] < siz[s])son[u] = s;}
}void dfs2(int u, int t)
{top[u] = t;if(son[u] == 0)return;dfs2(son[u], t);for(int i = 0; i < edge[u].size(); i ++){int s = edge[u][i].first;if(s == son[u] || s == fa[u])continue;dfs2(s, s);}
}int lca(int u, int v)
{while(top[u] != top[v]){if(dep[top[u]] < dep[top[v]])swap(u, v);u = fa[top[u]];}return dep[u] < dep[v] ? u : v;
}void cal_sum(int u)
{for(int i = 0; i < edge[u].size(); i ++){int s = edge[u][i].first;if(s == fa[u])continue;int w = edge[u][i].second;sum[s] = sum[u] + w;cal_sum(s);}
}void solve()
{cin >> n >> k;for(int i = 0; i < n - 1; i ++){int x, y, t;cin >> x >> y >> t;edge[x].push_back({y, t});edge[y].push_back({x, t});}for(int i = 1; i <= k; i ++)cin >> a[i];//树链剖分dfs1(1, 0);dfs2(1, 1);//求树上的前缀和cal_sum(1);int ans = 0;for(int i = 1; i <= k - 1; i ++){int u = a[i], v = a[i + 1];int cost = sum[u] + sum[v] - 2 * sum[lca(u, v)];ans += cost;}for(int i = 1; i <= k; i ++){int tmp = ans;if(i == 1)tmp -= sum[a[i + 1]] + sum[a[i]] - sum[lca(a[i], a[i + 1])] * 2;else if(i == k)tmp -= sum[a[i - 1]] + sum[a[i]] - sum[lca(a[i], a[i - 1])] * 2;else{tmp -= sum[a[i + 1]] + sum[a[i]] - sum[lca(a[i], a[i + 1])] * 2;tmp -= sum[a[i - 1]] + sum[a[i]] - sum[lca(a[i], a[i - 1])] * 2;tmp += sum[a[i - 1]] + sum[a[i + 1]] - sum[lca(a[i + 1], a[i - 1])] * 2;}cout << tmp << " ";}cout << endl;
}signed main()
{ios::sync_with_stdio(0);cin.tie(0);int t = 1;while(t--)solve();
}