90分WA#5#11求助！

我直接小根堆优化的dij+二分（找钱的方案），用的第二篇题解的思路

昨天调到今天力，有点崩溃了

#include<bits/stdc++.h>
#define ll long long
using namespace std;

const int N = 100005, M = 100005;
ll f[N], dis[N], n, m, b;
bool vis[N];
int head[N], ver[M], edge[M], nxt[M], num = 0; //Chain Forward Star

void add(int x, int y, int c) {
    ver[++num] = y, edge[num] = c;
    nxt[num] = head[x], head[x] = num;
}

priority_queue < pair<ll, int> > q;

bool chk(ll cst) {
    for(int i = 1; i <= n; i++) {
        dis[i] = LONG_LONG_MAX;
        vis[i] = 0; 
    }

    q.push(make_pair(0, 1));
    dis[1] = 0;
    if(f[1] > cst) return false;
    while(!q.empty()) {
        pair <ll, int> p = q.top();
        q.pop();
        ll dist = -p.first, x = p.second;

        if(x == n) {
            if(dis[n] > b) return false;
            return true;
        }
        if(vis[x]) continue;
        vis[x] = true;
        for(int i = head[x]; i > 0; i = nxt[i]) {
            ll y = ver[i], c = edge[i];
            if(dist + c <= dis[y] && f[y] <= cst)  {
                dis[y] = dist + c;
//              cout  << y << "》》"<< dis[y] << endl;
                q.push(make_pair(-dis[y], y));
            }
        }
    }
//  for(int i = 1; i <= n; i++)
//      cout << dis[i] << " ";
//  cout << endl;
//  if(dis[n] < cst) return true;
//  return false;
}

int main() {
//  freopen("a.aaa" , "w", stdout);
    cin >> n >> m >> b;
    for(int i = 1; i <= n; i++) cin >> f[i];

    int x, y, c;
    for(int i = 1; i <= m; i++) {
        cin >> x >> y >> c;
        add(x, y, c);
        add(y, x, c);
    }

    long long l = 1, r = 1000000005;
    if(!chk(r)) {
        cout << "AFK" << endl;
        return 0;
    }
    long long mid = (l + r) >> 1;
    while(l <= r) {
        bool tmp = chk(mid);
//      cout << l << " " << r << " " << tmp<< endl;
        if(tmp == 1) {
//          cout  << " Hello " << r << endl;
            r = mid - 1;
            mid = (l + r) >> 1;
        } else {
            l = mid + 1;
            mid = (l + r) >> 1;
        }
    } 
    cout << l << endl; 
    return 0;
}