[蓝桥杯]真题讲解：景区导游（DFS遍历、图的存储、树上前缀和与LCA）

一、视频讲解

视频讲解

二、暴力代码

//暴力代码：DFS
#include
#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
#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();
}