BZOJ3040: 最短路(road)

https://www.lydsy.com/JudgeOnline/problem.php?id=3040

  题目已经给出是最短路，所以我们可以用dijkstra（关于spfa它死了），但是看一眼这道题的数据范围，普通的dij肯定是会超时的。所以要优化（我不会优化所以直接借鉴了hzwer学长的代码）这里用的是平板电视优化QAQ

#include<cstdio>
#include<queue>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<ext/pb_ds/priority_queue.hpp>
#define llinf 9000000000000000000LL
typedef long long ll;
#define pa pair<ll,int>
using namespace std;
using namespace __gnu_pbds;
typedef __gnu_pbds::priority_queue<pa,greater<pa>,pairing_heap_tag > heap;
int n, m;
heap::point_iterator id[1000005];
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
struct edge {
    int to, nxt, from;
} e[10000005];
int tot;
int head[1000005];
ll dis[1000005];
void add(int x, int y, int z) {
    e[++tot].nxt = head[x];
    e[tot].to = y;
    e[tot].from = z;
    head[x] = tot;
}
void dijkstra() {
    heap q;
    for(int i = 1; i <= n; i++)
        dis[i] = llinf;
    dis[1] = 0;
    id[1] = q.push(make_pair(0,1));
    while(!q.empty()) {
        int x = q.top().second;
        q.pop();
        for(int i = head[x]; i; i = e[i].nxt)
            if(dis[e[i].to] > dis[x] + e[i].from) {
                dis[e[i].to] = dis[x] + e[i].from;
                if(id[e[i].to] != 0)
                    q.modify(id[e[i].to],make_pair(dis[e[i].to],e[i].to));
                else
                    id[e[i].to] = q.push(make_pair(dis[e[i].to],e[i].to));
            }
    }
}
int T, rxa, rxc, rya, ryc, rp, x, y, z;
int main() {
    n = read(), m = read();
    T = read(), rxa = read(), rxc = read(), rya = read(), ryc = read(), rp = read();
    int a, b;
    for(int i = 1; i <= T; i++) {
        x = ((ll)x * rxa + rxc) % rp;
        y = ((ll)y * rya + ryc) % rp;
        a = min(x % n + 1,y % n + 1);
        b = max(y % n + 1,y % n + 1);
        add(a,b,1e8 - 100 * a);
    }
    for(int i = 1; i <= m - T; i++) {
        x = read(), y = read(), z = read();
        add(x,y,z);
    }
    dijkstra();
    printf("%lld", dis[n]);
    return 0;
}