一個極其傻逼得重構樹，然而我數組開小了

#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
#include <cctype>
using namespace std;
const int N = 4e5 + 10, M = 1e6 + 10, inf = 0x7fffffff;
int v[M], to[M], head[N], pos[M], p;
void build(int a, int b, int c) { v[++ p] = b; to[p] = head[a]; head[a] = p; pos[p] = c; }
 
void link(int a, int b, int c) { build(a, b, c); build(b, a, c); }
int pre[N], far[N], n;
int find(int x) { return pre[x] == x ? x : pre[x] = find(pre[x]); }
struct point {
    int d, f;
    point (int d, int f) : d(d), f(f) {}
    point () {}
    bool operator < (const point &a) const {
        
 
return f > a.f;
    }
};
priority_queue<point> tp;
void dijkstra() {
    for ( int i = 2; i <= n; ++ i) far[i] = inf; tp.push(point(1, 0));
    while( !tp.empty()) {
        int u = tp.top().d; tp.pop();
        for ( int i = head[u]; i; i = to[i]) if( far[v[i]] > far[u] + pos[i]) {
            far[v[i]] 
 
= far[u] + pos[i]; tp.push(point(v[i], far[v[i]]));
        }
    }
}
int POS[N], T[N][30], cnt;
int query(int a, int P) {
    for ( int i = 20; i >= 0; -- i) if( POS[T[a][i]] > P)
        a = T[a][i];
    return far[a];
}
struct node {
    int u, v, val;
    node ( int u, int v, int val) : u(u), v(v), val(val) {}
    node () {}
    bool operator < (const node &a) const {
        return val > a.val;
    }
}tc[M];
int lastans, Case, m, q, K, s;
inline void G(int &x) {
    x = 0; char o = getchar(); int f = 1; for ( ; !isdigit(o); o = getchar()) if( o == ‘-‘) f = -1;
    for ( ; isdigit(o); o = getchar()) x = (x << 1) + (x << 3) + (o & 15); x *= f;
}
int TP[30];
int main() {
    freopen("return1.in", "r", stdin);
    freopen("return.out", "w", stdout);
    G(Case);
    while( Case -- ) {
        G(n), G(m);
        for ( int i = 1; i <= n; ++ i) head[i] = 0;
        for ( int i = 1, a, b, c, d; i <= m; ++ i) 
            G(a), G(b), G(c), G(d), tc[i] = node(a, b, d), link(a, b, c);
        sort(tc+1, tc+1+m); dijkstra();
        for ( int i = 1; i <= n; ++ i) pre[i] = i, POS[i] = inf;
        cnt = n;
        for ( int i = 1; i <= m; ++ i) {
            if( find(tc[i].u) == find(tc[i].v)) continue;
            int A = find(tc[i].u), B = find(tc[i].v);
            ++ cnt; POS[cnt] = tc[i].val; far[cnt] = min(far[A], far[B]);
            T[A][0] = T[B][0] = cnt;
            pre[cnt] = pre[A] = pre[B] = cnt;
        }
        T[cnt][0] = T[cnt][1] = cnt;
        for ( int i = 1; i <= 20; ++ i) 
            for ( int k = 1; k <= cnt; ++ k) T[k][i] = T[T[k][i-1]][i-1];
        G(q); G(K); G(s);
        cnt = 0;
        for ( int i = 1, v0, p0; i <= q; ++ i) {
            G(v0); G(p0);
            if( K) v0 = (v0 + K * lastans - 1) % n + 1, p0 = (p0 + K * lastans) % (s + 1);
            lastans = query(v0, p0); int t = lastans;
            do TP[++ cnt] = t % 10, t /= 10; while( t);
            for( ; cnt; -- cnt) putchar(‘0‘+TP[cnt]); puts("");
        }
    }
    return 0;
}

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