题目大意：给出每个点的权值和边，假设边为(u, v)，那么该边的权值就是(val[v] - val[u]) ^ 3
问从点1出发，到达所给点的最短路

解题思路：有可能会出现负环，出现负环的话，就将负环内的点所能到达的所有点都标记一下

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int N = 210;
const int M = 100010;
const int INF = 0x3f3f3f3f;
struct Edge{
    int u, v, dis, next;
    Edge() {}
    Edge(int u, int v, int dis, int next): u(u), v(v), dis(dis), next(next) {}
}E[M * 2];

int val[N], head[N] ,cnt[N], d[N];
bool vis[N], cycle[N];
int n, m, cas = 1;

void dfs(int u) {
    cycle[u] = true;
    for (int i = head[u]; ~i; i = E[i].next){
        int v = E[i].v;
        if (cycle[v]) continue;
        dfs(v);
    }
} 

void solve() {
    queue<int> Q;
    for (int i = 1; i <= n; i++) d[i] = INF;
    d[1] = 0;
    memset(vis, 0, sizeof(vis));
    memset(cnt, 0, sizeof(cnt));
    memset(cycle, 0, sizeof(cycle));
    Q.push(1);
    while (!Q.empty()) {
        int u = Q.front(); Q.pop();
        vis[u] = false;

        for (int i = head[u]; ~i; i = E[i].next) {
            int v = E[i].v;
            if (cycle[v]) continue;
            if (d[v] > d[u] + E[i].dis) {
                d[v] = d[u] + E[i].dis;
                if (!vis[v]) {
                    vis[v] = true;
                    Q.push(v);
                    if (++cnt[v] >= n) {
                        dfs(v);
                    }
                }
            }
        }
    }

    int q, u;
    printf("Case %d:\n", cas++);
    scanf("%d", &q);
    while (q--) {
        scanf("%d", &u);

        if (cycle[u] || d[u] < 3 || d[u] == INF) printf("?\n");
        else printf("%d\n", d[u]);
    }
}

int getdis(int t) {
    return t * t * t;
}

void init() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        scanf("%d", &val[i]);

    memset(head, -1, sizeof(head));
    scanf("%d", &m);

    int u, v;
    for (int i = 0; i < m; i++) {
        scanf("%d%d", &u, &v);
        E[i] = Edge(u, v, getdis(val[v] - val[u]), head[u]);
        head[u] = i;
    }
}

int main() {
    int test;
    scanf("%d", &test);
    while (test--) {
        init();
        solve();
    }
    return 0;
}