hdu 2586 How far away ？（LCA离线）_it uuuuuu okipiilkv-优快云博客

本文链接：https://blog.youkuaiyun.com/u014204835/article/details/41986213

本文介绍了一种基于Tarjan算法实现的路径查询方法。通过建立深搜树，该方法可以高效地解决图中节点间的距离计算问题。文章详细展示了算法的实现过程，并通过具体代码解释了如何进行初始化、边的添加、搜索树构建等关键步骤。

记录下所有的提问，在建tarjan建一颗深搜树的时候询问。奇怪这道题的edge开4W会RE，，于是任性的开了40W妥妥过.

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>

using namespace std;

const int N = 400005;
const int M = 205;

struct node{
	int v, w, nxt;
}e[N];

struct pp{
	int v, id;
}p;

vector <pp> ask[N];
int head[N];
int cnt;
int vis[N];
int ans[N];
int n, m;
int pre[N];
int dis[N];

void add(int u, int v, int w)
{
	e[cnt].w = w;
	e[cnt].v = v;
	e[cnt].nxt = head[u];
	head[u] = cnt++;
}

void init()
{
	memset(vis, 0, sizeof(vis));
	cnt = 0;
	memset(head, -1, sizeof(head));
	for( int i = 1; i <= n; i ++ )
		ask[i].clear();
	for( int i = 1; i <= n; i++ )
		pre[i] = i;
	memset(dis, 0, sizeof(dis));
	memset(ans, -1, sizeof(ans));
}

int find(int x)
{
	return x == pre[x] ? x : pre[x] = find(pre[x]);
}

void tarjan(int u)
{
	for( int i = head[u]; ~i; i = e[i].nxt )
	{
		int to = e[i].v;
		if( !vis[to] )
		{
			vis[to] = 1;
			dis[to] = dis[u] + e[i].w;
			tarjan(to);
			pre[to] = u;
			for( int j = 0; j < ask[to].size(); j++ )
			{
				int v = ask[to][j].v;
				if( vis[v] && ans[ask[to][j].id] == -1 )
				{
					if( v == to )
					{
						ans[ask[to][j].id] = 0;
					}
					else
					{
						ans[ask[to][j].id] = dis[to] + dis[v] - 2 * dis[find(v)];
					}
				}
			}
		}
	}
}

int main()
{
	int tot;
	scanf("%d", &tot);
	while(tot--)
	{
		scanf("%d%d", &n, &m);
		init();
		int u, v, w;
		for(int i = 1; i < n; i ++ )
		{
			scanf("%d%d%d", &u, &v, &w);
			add(u, v, w);
			add(v, u, w);
		}
		for( int i = 1; i <= m; i ++ )
		{
			scanf("%d%d", &u, &v);
			p.v = v, p.id = i;
			ask[u].push_back(p);
			p.v = u;
			ask[v].push_back(p);
		}
		vis[1] = 1;
		tarjan(1);
		for( int i = 1; i <= m; i ++ )
			printf("%d\n", ans[i]);
	}
	return 0;
}