luoguP3359 改造异或树

https://www.luogu.org/problemnew/show/P3359

因为 a ^ b ^ b = a，所以我们预处理 1 到所有点的距离，将删边的操作反过来变成加边，对于每一个联通块用 map 维护 1 到联通块中的点异或值为 x 的数的个数，乘法原理统计答案，加边时启发式合并即可

#include <bits/stdc++.h>

using namespace std;

typedef unsigned long long ull;

typedef long long ll;

template <typename T>

inline void read(T &f) {

	f = 0; T fu = 1; char c = getchar();

	while(c < '0' || c > '9') {if(c == '-') fu = -1; c = getchar();}

	while(c >= '0' && c <= '9') {f = (f << 3) + (f << 1) + (c & 15); c = getchar();}

	f *= fu;

}

const int N = 1e5 + 5;

struct Edge {

	int u, v, next, val;

}G[N << 1];

map <int, int> t[N];

ll Ans[N], ans;

int d[N], f[N], del[N], x[N], y[N], z[N], head[N];

int n, tot;

inline void addedge(int u, int v, int val) {

	G[++tot] = (Edge) {u, v, head[u], val}, head[u] = tot;

	G[++tot] = (Edge) {v, u, head[v], val}, head[v] = tot;

}

int find(int x) {return f[x] == x ? x : f[x] = find(f[x]);}

void dfs(int u, int fa) {

	for(int i = head[u]; i; i = G[i].next) {

		int v = G[i].v;

		if(v == fa) continue;

		d[v] = d[u] ^ G[i].val;

		dfs(v, u);

	}

}

int main() {

	cin >> n;

	for(int i = 1; i < n; i++) {

		read(x[i]); read(y[i]); read(z[i]);

		addedge(x[i], y[i], z[i]);

	}

	for(int i = 1; i < n; i++) read(del[i]);

	for(int i = 1; i <= n; i++) f[i] = i;

	dfs(1, 0); for(int i = 1; i <= n; i++) t[i][d[i]] = 1;

	for(int i = n - 1; i >= 1; i--) {

		int l = x[del[i]], r = y[del[i]];

		int x = find(l), y = find(r);

		if(t[x].size() > t[y].size()) swap(x, y);

		for(map <int, int> :: iterator it = t[x].begin(); it != t[x].end(); it++) {

			ans += (ll)it -> second * (ll)t[y][it -> first];

			t[y][it -> first] += it -> second;

		}

		f[x] = y; Ans[i] = ans; t[x].clear();

	}

	for(int i = 1; i <= n; i++) printf("%lld\n", Ans[i]);

	return 0;

}

巴特西

luoguP3359 改造异或树

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