题目意思

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define mk(__x__,__y__) make_pair(__x__,__y__)
#define ms(__x__,__val__) memset(__x__, __val__, sizeof(__x__))
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }
const int dir[][2] = { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;

const int N = 800 + 7; //节点数
const int M = 4000 + 7; //路径数
int head[N], tot = 0;//前向星变量
struct Node {
    //int u; //起点
    int w; //权值
    int v, next;
} edge[M << 1];

struct Temp {
    int u, k;
    ll dis;
    bool operator < (const Temp& A) const {
        return dis > A.dis;
    }
};

void add(int u, int v, int w) {
    tot++;
    //edge[tot].u = u;
    edge[tot].v = v;
    edge[tot].w = w;
    edge[tot].next = head[u];
    head[u] = tot;
}

int a[N];
ll d[N][11];
bool vis[N][11];
int n, m, k;

void dijkstra(int st, int ed) {
    ms(vis, 0);    ms(d, 0x3f);
    d[st][0] = 0;
    priority_queue<Temp> pq;
    pq.push({ st,0,0 });
    while (pq.size()) {
        Temp now = pq.top(); pq.pop();
        if (vis[now.u][now.k])    continue;
        vis[now.u][now.k] = 1;
        for (int i = head[now.u]; i; i = edge[i].next) {
            int v = edge[i].v;
            int w = edge[i].w;
            int tk = a[now.u] + now.k;
            if (tk <= k and !vis[v][tk] and d[v][tk] > now.dis + w) {
                d[v][tk] = now.dis + w;
                pq.push({ v,tk,d[v][tk] });
            }
        }
    }
}

int main() {
    n = read(), m = read(), k = read();
    for (int i = 1; i <= n; ++i)    a[i] = read();
    for (int i = 1; i <= m; ++i) {
        int u = read(), v = read(), w = read();
        add(u, v, w), add(v, u, w);
    }
    dijkstra(1, n);
    ll ans = INF;
    for (int i = 0; i <= k; ++i)    ans = min(ans, d[n][i]);
    if (ans == INF)    puts("-1");
    else    print(ans);
    return 0;
}