为什么要把魔法值的上限设置为 100，不能设置为 max(w[i])

#include <bits/stdc++.h>
#define For(i, a, b) for (int i = (a); i <= (b); i++)
#define pb push_back
#define mem(a, b) memset((a), (b), sizeof(a))
// #define int long long
#define endl "\n"
#define io                        \
    std::ios::sync_with_stdio(0); \
    std::cin.tie(0);              \
    std::cout.tie(0);

using namespace std;
typedef pair<int, int> PII;
typedef unsigned long long u64;

// long long long
// 数组开够 N+10
// 有向图 or 无向图
// 多组数据,初始化
// LLLLLLLLLLLLL
const int N = 1e5 + 10, M = 1e3 + 10;
const int inf = 1e9;
const int mod = 1e9 + 7;

int n, m, idx, a, b, c, k, res;
int h[N], e[N], ne[N], de[N], w[N];
int dis[N], st[105][N];
int dp[105][N];

void add(int a, int b) {
    e[idx] = b;
    ne[idx] = h[a];
    h[a] = idx++;
}

void add(int a, int b, int c) {
    e[idx] = b;
    w[idx] = c;
    ne[idx] = h[a];
    h[a] = idx++;
}
int s, t;
typedef pair<int, PII> my;

signed main() {
    io;
    mem(h, -1);
    mem(dp, 0x3f);

    cin >> n >> m >> t >> s;
    int mx = -1;

    For(i, 1, m) {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
        add(b, a, c);
        mx = max(mx, c);
    }

    priority_queue<my, vector<my>, greater<my>> heap;
    heap.push({0, {0, s}});
    dp[0][s] = 0;
    // cout << dp[0][s] << endl;

    // 这一行 mx=100 如果注释了
    // 就会 wa
    mx = 100;

    // cout << mx << endl;

    while (heap.size()) {
        auto p = heap.top();
        heap.pop();

        auto pp = p.second;
        auto mp = pp.first;
        auto x = pp.second;

        if (st[mp][x])
            continue;
        st[mp][x] = 1;

        if (mp > mx + 1)
            continue;

        auto d = dp[mp][x];
        // cout << x << " " << mp << " " << d << endl;
        for (int i = h[x]; i != -1; i = ne[i]) {
            int j = e[i];
            int to = mp + 1;
            int cost = w[i] / (to);
            if (dp[to][j] > d + cost) {
                dp[to][j] = d + cost;
                heap.push({dp[to][j], {to, j}});
            }
        }
    }
    res = inf;

    For(i, 0, mx + 1) {
        res = min(res, dp[i][t]);
    }
    cout << res;

    return 0;
}