CodeForces 464E The Classic Problem | Dexer chair tree maintains high precision

topic description

There is an undirected graph with \(n\) points \(m\) edges, and the edge weight of the \(i\)th edge is \(2^{a_i}\) . Find the shortest path length from point \(s\) to point \(t\) ( modulo \(10^9 + 7\) ).

answer

The idea is very simple - maintain the \(dis\) of each point with the chairman tree . Because every time the \(dis_v\) of a certain point \(v\) is updated , the new \(dis_v\) is the \ (dis_u + 2^{w_{u, v}}\) , which is equivalent to modifying the chair tree corresponding to the original \(u\) to obtain a new chair tree as the chair tree corresponding to \(v\) .

How to maintain the binary high precision of the chairman tree (line segment tree)? It's good to maintain it like loose loose = =

Questions that require (wǒ) to (fàn) note (guò) meaning:

1. If you use it priority_queueto do Dijkstra, and modify the dis corresponding to the node in the middle, it will affect the nature of the heap, and it will be WA. The correct way is to priority_queuepass pair<节点编号，当前dis>.
2. Should the space of the chairman tree be properly optimized? For example, during a query operation, pushdown will create new nodes, but these nodes will not be used in the future. Therefore, after a complete query operation, these new nodes can be deleted, and the space can be significantly optimized.

code

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#define space putchar(' ')
#define enter putchar('\n')
typedef long long ll;
using namespace std;
template <class T>
void read(T &x){
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
        if(c == '-') op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9')
        x = x * 10 + c - '0';
    if(op) x = -x;
}
template <class T>
void write(T x){
    if(x < 0) putchar('-'), x = -x;
    if(x >= 10) write(x / 10);
    putchar('0' + x % 10);
}

const int N = 100100, M = 40000007, mod = 1000000007, P = 1000000021;
int n, s, t, maxn = 100098, m, hsh100[N], hsh111[N], ans100[N], ans111[N];
int ecnt, adj[N], nxt[2*N], go[2*N], w[2*N], pre[N], stk[N], top;
int ls[M], rs[M], hsh[M], ans[M], tot, root[N];
bool lazy[M], vis[N];
void adde(int u, int v, int ww){
    go[++ecnt] = v;
    nxt[ecnt] = adj[u];
    adj[u] = ecnt;
    w[ecnt] = ww;
}
int newnode(int old){
    int k = ++tot;
    ls[k] = ls[old], rs[k] = rs[old];
    hsh[k] = hsh[old], ans[k] = ans[old], lazy[k] = lazy[old];
    return k;
}
int pushdown(int k){
    if(!lazy[k]) return newnode(k);
    k = newnode(k);
    lazy[k] = 0;
    ls[k] = newnode(ls[k]), rs[k] = newnode(rs[k]);
    lazy[ls[k]] = lazy[rs[k]] = 1;
    hsh[ls[k]] = hsh[rs[k]] = ans[ls[k]] = ans[rs[k]] = 0;
    return k;
}
int change0(int k, int l, int r, int ql, int qr){
    if(ql <= l && qr >= r) return k = newnode(k), lazy[k] = 1, hsh[k] = ans[k] = 0, k;
    k = pushdown(k);
    int mid = (l + r) >> 1;
    if(ql <= mid) ls[k] = change0(ls[k], l, mid, ql, qr);
    if(qr > mid) rs[k] = change0(rs[k], mid + 1, r, ql, qr);
    hsh[k] = (hsh[ls[k]] + (ll)hsh[rs[k]] * hsh100[mid - l + 1]) % P;
    ans[k] = (ans[ls[k]] + (ll)ans[rs[k]] * ans100[mid - l + 1]) % mod;
    return k;
}
int change1(int k, int l, int r, int p){
    if(l == r) return k = newnode(k), lazy[k] = 0, hsh[k] = ans[k] = 1, k;
    k = pushdown(k);
    int mid = (l + r) >> 1;
    if(p <= mid) ls[k] = change1(ls[k], l, mid, p);
    else rs[k] = change1(rs[k], mid + 1, r, p);
    hsh[k] = (hsh[ls[k]] + (ll)hsh[rs[k]] * hsh100[mid - l + 1]) % P;
    ans[k] = (ans[ls[k]] + (ll)ans[rs[k]] * ans100[mid - l + 1]) % mod;
    return k;
}
int find0(int k, int l, int r, int ql, int qr){
    if(hsh[k] == hsh111[r - l + 1] && ans[k] == ans111[r - l + 1]) return -1;
    if(l == r) return l;
    k = pushdown(k);
    int mid = (l + r) >> 1;
    if(ql > mid) return find0(rs[k], mid + 1, r, ql, qr);
    int ret = find0(ls[k], l, mid, ql, qr);
    if(ret != -1) return ret;
    return find0(rs[k], mid + 1, r, ql, qr);
}
int add(int k, int p){
    int mem_tot = tot;
    int q = find0(k, 0, maxn, p, maxn);
    tot = mem_tot;
    k = change1(k, 0, maxn, q);
    if(p < q) k = change0(k, 0, maxn, p, q - 1);
    return k;
}
bool diff(int k1, int k2, int l, int r){
    if(l == r) return hsh[k1] < hsh[k2];
    k1 = pushdown(k1), k2 = pushdown(k2);
    int mid = (l + r) >> 1;
    if(hsh[rs[k1]] == hsh[rs[k2]] && ans[rs[k1]] == ans[rs[k2]])
        return diff(ls[k1], ls[k2], l, mid);
    else return diff(rs[k1], rs[k2], mid + 1, r);
}
struct Data {
    int node, root;
    bool operator < (const Data &b) const {
        int mem_tot = tot;
        bool ret = diff(b.root, root, 0, maxn);
        tot = mem_tot;
        return ret;
    }
};
priority_queue <Data> que;

int main(){

    read(n), read(m);
    hsh100[0] = ans100[0] = 1;
    for(int i = 1; i <= maxn + 1; i++){
        hsh100[i] = hsh100[i - 1] * 2 % P;
        ans100[i] = ans100[i - 1] * 2 % mod;
        hsh111[i] = (hsh100[i] - 1 + P) % P;
        ans111[i] = (ans100[i] - 1 + mod) % mod;
    }
    for(int i = 1, u, v, ww; i <= m; i++)
        read(u), read(v), read(ww), adde(u, v, ww), adde(v, u, ww);
    read(s), read(t);
    root[0] = add(0, maxn - 1);
    for(int i = 1; i <= n; i++)
        if(i != s) root[i] = root[0];
    que.push((Data){s, root[s]});
    while(!que.empty()){
        int u = que.top().node;
        que.pop();
        if(vis[u]) continue;
        vis[u] = 1;
        for(int e = adj[u], v; e; e = nxt[e]){
            v = go[e];
            int tmp = add(root[u], w[e]);
            if(diff(tmp, root[v], 0, maxn))
                root[v] = tmp, pre[v] = u, que.push((Data){v, root[v]});
        }
    }
    if(ans[root[t]] == ans100[maxn - 1] && hsh[root[t]] == hsh100[maxn - 1])
        return puts("-1"), 0;
    write(ans[root[t]]), enter;
    stk[++top] = t;
    while(pre[stk[top]]) stk[top + 1] = pre[stk[top]], top++;
    write(top), enter;
    while(top) write(stk[top--]), top ? space : enter;

    return 0;
}