CF938G Shortest Path Queries

Solution

套路题。

$xor$最短路可以用线性基维护（把每个环的边权异或和放进线性基，询问时把树边路径边权异或和放在线性基里跑出最小值即可）。

然后因为线性基删除比较慢而麻烦（注意线性基是有办法动态加删元素的），因此动态加删边可以用线段树分治+可撤销带权并查集实现。

时间复杂度$O(nl{g}^{2}n)$。

Code（最近顺便改了改码风）

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std;template<typename T> inline bool upmin(T &x, T y) { return y < x ? x = y, 1 : 0; }
template<typename T> inline bool upmax(T &x, T y) { return x < y ? x = y, 1 : 0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int, int> PR;
typedef vector<int> VI;const lod eps = 1e-11;
const lod pi = acos(-1);
const int oo = 1 << 30;
const ll loo = 1ll << 62;
const int mods = 998244353;
const int MAXN = 1000005;
const int INF = 0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f = 1, x = 0; char c = getchar();while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar(); }while (c >= '0' && c <= '9') { x = (x << 3) + (x << 1) + (c ^ 48); c = getchar(); }return x * f;
}
map<PR, PR> Map; 
int n, m, base[30], btop = 0, bstk[MAXN], f[MAXN], num[MAXN], d[MAXN], Ans[MAXN], utop = 0;
struct ustknode{ int u, v, c; } ustk[MAXN];
struct Cnode{ int x, y, c; };
struct Qnode{ int id, x, y; };
vector<Cnode> C[MAXN];
vector<Qnode> Q[MAXN];  void insert(int x) {for (int i = 29; i >= 0; -- i)if ((x >> i) & 1) {if (!base[i]) { base[i] = x, bstk[++btop] = i; return; }x ^= base[i];}
}
int getmn(int x) {for (int i = 29; i >= 0; -- i) upmin(x, x ^ base[i]);return x;
} int find(int x) { return f[x] == x ? f[x] : find(f[x]); }
int getdis(int x) { return f[x] == x ? d[x] : getdis(f[x]) ^ d[x]; }
void unions(int u, int v, int c) {if (num[u] < num[v]) swap(u, v);ustk[++utop] = (ustknode){u, v, c}, f[v] = u, d[v] ^= c, num[u] += num[v]; 
}
void update(int x, int l, int r, int L, int R, Cnode y)
{if (l >= L && r <= R) { C[x].PB(y); return; }int mid = (l + r) >> 1;if (R <= mid) update(x << 1, l, mid, L, R, y);else if (L > mid) update(x << 1 | 1, mid + 1, r, L, R, y);else update(x << 1, l, mid, L, mid, y), update(x << 1 | 1, mid + 1, r, mid + 1, R, y);
}
void solve(int x, int l, int r) {int blst = btop, ulst = utop;for (auto t:C[x]) {int u = t.x, v = t.y, c = t.c, uu = find(u), vv = find(v);if (uu == vv) insert(getdis(u) ^ getdis(v) ^ c);else unions(uu, vv, getdis(u) ^ getdis(v) ^ c);  }if (l == r) {for (auto t:Q[l]) Ans[t.id] = getmn(getdis(t.x) ^ getdis(t.y));}else {int mid = (l + r) >> 1;solve(x << 1, l, mid);solve(x << 1 | 1, mid + 1, r);}while (btop > blst) base[bstk[btop --]] = 0;while (utop > ulst) num[ustk[utop].u] -= num[ustk[utop].v], f[ustk[utop].v] = ustk[utop].v, d[ustk[utop].v] ^= ustk[utop].c, -- utop;
}
signed main() {n = read(), m = read();for (int i = 1; i <= n ; ++ i) num[i] = 1, f[i] = i;for (int i = 1; i <= m ; ++ i) {int u = read(), v = read(), c = read();Map[MP(u, v)] = MP(0, c);}int Case = read(), num = 0;for (int i = 1; i <= Case ; ++ i) {int opt = read(), x = read(), y = read(), c;if (opt == 1) c = read(), Map[MP(x, y)] = MP(i, c);if (opt == 2) update(1, 0, m, Map[MP(x, y)].fi, i, (Cnode){x, y, Map[MP(x,y)].se}), Map.erase(MP(x, y));if (opt == 3) Q[i].PB(((Qnode){++ num, x, y} ));}for (auto t : Map) update(1, 0, m, t.se.fi, m, (Cnode){t.fi.fi, t.fi.se, t.se.se});solve(1, 0, m);for (int i = 1; i <= num; ++ i) printf("%d\n", Ans[i]);return 0;
}