题意描述
有一个\(n\)点\(m\)边的无向图，第\(i\)条边的边权是\(2^{a_i}\)。求点\(s\)到点\(t\)的最短路长度（对\(10^9 + 7\)取模）。
题解
思路很简单——用主席树维护每个点的\(dis\)。因为每次更新某个点\(v\)的\(dis_v\)的时候，新的\(dis_v\)都是某个点\(u\)的\(dis_u + 2^{w_{u, v}}\)，相当于在原先\(u\)对应的主席树基础上修改，得到新的一棵主席树，作为\(v\)对应的主席树。
主席树（线段树）维护二进制高精度怎么维护呢？像松松松那么维护就好了 = =
需（wǒ）要（fàn）注（guò）意的问题：
- 如果你用
priority_queue来做Dijkstra，又中途修改了节点对应的dis，会影响堆的性质，会WA。正确做法是在priority_queue传pair<节点编号，当前dis>。 - 主席树的空间要适当优化优化？例如查询操作的时候，pushdown会创造新的节点，但是以后就不会用到这群节点了，于是一次完整的查询操作之后把这些新节点都删掉就好了，空间可以得到明显优化。
代码
#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;
}原文地址：https://www.cnblogs.com/RabbitHu/p/CF464E.html
时间： 2024-10-13 08:08:42