How far away ？ HDU - 2586 LCA的Tarjan算法-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_43326028/article/details/95446212

博客对离线LCA(Tarjan)算法进行了详细讲解，聚焦于信息技术领域的算法知识。

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#include<iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<vector>
#include<cmath>
#define LL long long 
#define mem(a, b) memset(a, b, sizeof(a))
#define N 50010 
#define MOD
using namespace std;
int ver[N*2], Next[2*N], edge[N*2], head[N];
int fa[N], d[N], v[N], lca[N], ans[N];
vector<int> query[N], query_id[N];
int T, n, m, tot, t;

void add(int x, int y, int z) {//邻接表 
	ver[++tot] = y, edge[tot] = z, Next[tot] = head[x], head[x] = tot;
} 

void add_query(int x, int y, int id) {//离线纪律每个询问，记录询问x,y，记录是第几次询问 
	query[x].push_back(y), query_id[x].push_back(id);
	query[y].push_back(x), query_id[y].push_back(id);
}

int get(int x) {//并查集找父节点 
	if (x == fa[x]) return x;
	return fa[x] = get(fa[x]);
}

void tarjan(int x) {
	v[x] = 1;
	for (int i = head[x]; i; i = Next[i]) {
		int y = ver[i];//子节点 
		if (v[y]) continue;//如果该子节点被访问过，就不访问 
		d[y] = d[x] + edge[i];//更新距离 
		tarjan(y);//往下访问该节点的子节点 
		fa[y] = x;//记录y的父节点是x 
	}
	for (int i = 0; i < query[x].size(); i++) {//如果有关于当前节点的询问 
		int y = query[x][i], id = query_id[x][i];//询问x与y之间的距离 
		if (v[y] == 2) {//如果y已经回溯 
			int lca = get(y);//找到y当前的父节点，也就是他们的LCA 
			ans[id] = min(ans[id], d[x] + d[y] - 2*d[lca]);
		}
	}
	v[x] = 2;//把v[x]的值该为2，表示该节点访问完毕回溯
} 

int main() {
	scanf("%d",&T);
	while(T--) {
		scanf("%d%d", &n, &m);
		for (int i = 1; i <= n; i++) {
			head[i] = 0, fa[i] = i, v[i] = 0;
			query[i].clear(), query_id[i].clear();
		}
		tot = 0;
		for (int i = 1; i < n; i++) {
			int x, y, z;
			scanf("%d%d%d", &x, &y, &z);
			add(x, y, z), add(y, x, z);
		} 
		for (int i = 1; i <= m; i++) {
			int x, y;
			scanf("%d%d", &x, &y);
			if(x == y) ans[i] = 0;
			else {
				add_query(x, y, i);
				ans[i] = 1 << 30;
			}
		}
		tarjan(1);
		for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
	} 
}