其他的部分和y总一样，就是最后求最小值的时候用的是stl的优先队列

之前写的代码不知道为啥一直29，所以用y总的代码改了一下

#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

typedef long long LL;
typedef pair<LL, int> PII;

const int N = 100010, M = 200010 * 2;
const LL INF = 0x3f3f3f3f3f3f3f3fll;

int n, m, Q;
int h1[N], h2[N], e[M], w[M], ne[M], idx;
LL dist1[N], dist2[N];
bool st[N];
int ratio[N];
LL g[N];
void add(int h[], int a, int b, int c)  
{
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++ ;
}

void dijkstra(int h[], LL dist[], int start)  
{
    memset(dist, 0x3f, sizeof dist1);
    memset(st, 0, sizeof st);
    dist[start] = 0;
    priority_queue<PII, vector<PII>, greater<PII>> heap;
    heap.push({0, start});

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

        int ver = t.second;

        if (st[ver]) continue;
        st[ver] = true;

        for (int i = h[ver]; i != -1; i = ne[i])
        {
            int j = e[i];
            if (dist[j] > dist[ver] + w[i])
            {
                dist[j] = dist[ver] + w[i];
                heap.push({dist[j], j});
            }
        }
    }
}

int main()
{
    scanf("%d%d%d", &n, &m, &Q);
    memset(h1, -1, sizeof h1);
    memset(h2, -1, sizeof h2);
    while (m -- )
    {
        int a, b, c, d;
        scanf("%d%d%d%d", &a, &b, &c, &d);
        add(h1, a, b, c), add(h2, b, a, d);
    }
    for (int i = 1; i <= n; i ++ ) scanf("%d", &ratio[i]);

    dijkstra(h1, dist1, 1);
    dijkstra(h2, dist2, n);
    priority_queue<PII, vector<PII>, greater<PII> > heap;
    for (int i = 1; i <= n; i ++ )
        if (dist1[i] != INF && dist2[i] != INF)
        {
            g[i] = dist1[i] + (dist2[i] + ratio[i] - 1) / ratio[i];
            //g[i]数组是用来存在i个点金币转换的最小消费的当前值
            heap.push({g[i],i});
        }

    while (Q -- )
    {
        int a, b;
        scanf("%d%d", &a, &b);

        if (dist1[a] != INF && dist2[a] != INF)
        {
            ratio[a] = b;
            LL t1 =dist1[a] + (dist2[a] + ratio[a] - 1) / ratio[a];
            g[a] = t1;
            heap.push({g[a],a});
        }
        PII t = heap.top();
        while (g[t.second] != t.first)
        //如果heap里面存的pair对的第二维值对应的g[]当前值和第一维值不等，
        //那么这个键值对不符合要求，因为值被更新了
        //当一些点的g[]被修改了，我们不好直接从优先队列中删除，
        //所以每次判断队头是否和当前g[]相同，不相同则应该删除
        {
            heap.pop();
            t=heap.top();
        }
        if (Q!=0)
        cout<<t.first<<endl;
        else cout<<t.first;
    }

    return 0;
}