思路：记录每个点与其根结点的横向距离和纵向距离，当知道其父节点与根结点的关系，很容易推出当前节点与根结点的关系：

直接相加即可。

int p = a[x].par;
a[x].dx += a[p].dx;
a[x].dy += a[p].dy;

合并什么的就不多说了，很容易得出。

值得一说的就是需要按照合并节点的顺序把询问排序，最后按照询问的顺序再排一次序。

AC代码

#include <cstdio>
#include <cmath>
#include <cctype>
#include <algorithm>
#include <cstring>
#include <utility>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define eps 1e-10
#define inf 0x3f3f3f3f
#define PI pair<int, int>
typedef long long LL;
const int maxn = 40000 + 5;
struct node{
	int par;
	int dx, dy;
}a[maxn];

struct Edge{
	int u, v, dis, ch;
}b[maxn];

struct Question{
	int u, v, t, num;
}q[10000+5];

void init(int n) {
	for(int i = 1; i <= n; ++i) {
		a[i].par = i;
		a[i].dx = a[i].dy = 0;
	}
}

int find(int x) {
	if(a[x].par == x) return x;
	int r = find(a[x].par);
	int p = a[x].par;
	a[x].dx += a[p].dx;
	a[x].dy += a[p].dy;
	return a[x].par = r;
}

void unionset(int x, int y, int dx, int dy) {
	int rx = find(x), ry = find(y);
	if(rx != ry) {
		a[rx].par = y;
		a[rx].dx = dx - a[x].dx;
		a[rx].dy = dy - a[x].dy;
	}
}

int get_dis(int x, int y) {
	int rx = find(x), ry = find(y);
	if(rx != ry) return -1;
	return abs(a[x].dx - a[y].dx) + abs(a[x].dy - a[y].dy);
}

bool cmp1(const Question &a, const Question &b) {
	return a.t < b.t;
}

bool cmp2(const PI &a, const PI &b) {
	return a.second < b.second;
}

int main() {
	int n, m, Q;
	while(scanf("%d%d", &n, &m) == 2) {
		init(n);
		for(int i = 1; i <= m; ++i) {
			scanf("%d %d %d %c", &b[i].u, &b[i].v, &b[i].dis, &b[i].ch);
			//printf("%d %d %d %c\n", b[i].u, b[i].v, b[i].dis, b[i].ch);
		}
		scanf("%d", &Q);
		for(int i = 1; i <= Q; ++i) {
			scanf("%d%d%d", &q[i].u, &q[i].v, &q[i].t);
			q[i].num = i;
		}
		vector<PI>ans;
		sort(q+1, q+Q+1, cmp1);
		int ind = 1;
		for(int i = 1; i <= m; ++i) {
			int dx, dy;
			switch(b[i].ch) {
				case 'E': dx = b[i].dis, dy = 0; break;
				case 'W': dx = -b[i].dis, dy = 0; break;
				case 'N': dx = 0, dy = b[i].dis; break;
				case 'S': dx = 0, dy = -b[i].dis; break;
			}
			unionset(b[i].u, b[i].v, dx, dy);
			while(ind <= Q && q[ind].t == i) {
				ans.push_back(make_pair(get_dis(q[ind].u, q[ind].v), q[ind].num));
				++ind;
			}
		}
		sort(ans.begin(), ans.end(), cmp2);
		for(int i = 0; i < ans.size(); ++i) printf("%d\n", ans[i].first);
	}
	return 0;
}

如有不当之处欢迎指出！