（带权并查集理解不通的我觉得可以用这题磨练一下）

重点：用dx[], dy[]去维护各个点的坐标的值

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;

const int N = 100010;

struct node
{
    int a, b, l;
    char d;
}s[N];

struct nodee
{
    int a, b, I, rank, ans; // rank记录原先的位置
}q[N];

int p[N], dx[N], dy[N];
int n, m, k;

bool cmp(nodee x, nodee y)
{
    return x.I < y.I;
}

bool cmpp(nodee x, nodee y) // 用于还原输出顺序
{
    return x.rank < y.rank;
}

void init()
{
    for (int i = 0; i < N; i ++)
        p[i] = i;
}

int find(int x)
{
    if (x != p[x])
    {
        int t = find(p[x]);
        dx[x] += dx[p[x]];
        dy[x] += dy[p[x]];
        p[x] = t;
    }
    return p[x];
}

int main()
{
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    init();
    cin >> n >> m;

    for (int i = 1; i <= m; i ++)
    {
        int a, b, l;
        char d;
        cin >> a >> b >> l >> d;
        s[i] = {a, b, l, d};
    }

    cin >> k;
    for (int i = 1; i <= k; i ++)
    {
        int a, b, I;
        cin >> a >> b >> I;
        q[i] = {a, b, I, i};
    }

    sort(q + 1, q + k + 1, cmp);

    int idx = 1;
    for (int i = 1; i <= m; i ++)
    {
        int a = s[i].a, b = s[i].b, l = s[i].l;
        char ch = s[i].d;

        int fa = find(a), fb = find(b);

        if (fa != fb)
        {
            p[fa] = fb;
            if (ch == 'E') dx[fa] = dx[b] - dx[a] - l, dy[fa] = dy[b] - dy[a]; // 这里调了两小时，也可能脑子真的混乱了。
            else if (ch == 'W') dx[fa] = dx[b] - dx[a] + l, dy[fa] = dy[b] - dy[a];
            else if (ch == 'N') dy[fa] = dy[b] - dy[a] - l, dx[fa] = dx[b] - dx[a];
            else dy[fa] = dy[b] - dy[a] + l, dx[fa] = dx[b] - dx[a];
        }

        while (idx <= k && i == q[idx].I) // 到哪个点存哪个值
        {
            int a = q[idx].a, b = q[idx].b;
            int fa = find(a), fb = find(b);

            if (fa != fb) q[idx].ans = -1;
            else q[idx].ans = abs(dx[a] - dx[b]) + abs(dy[a] - dy[b]);

            idx ++;
        }
    }

    sort(q + 1, q + k + 1, cmpp); // 记得将顺序返回（输出要求）

    for (int i = 1; i <= k; i ++) cout << q[i].ans << '\n';

    return 0;
}