洛谷 P5651 【基礎最短路練習題】

P5651

題目大意

求一個x到y簡單路徑的異或和.

思路：

因爲圖中肯定有環我們肯定是不能走環的.

然後我就隨便找了一顆生成樹，然後在生成樹上找到每一個點到根節點
異或和，查詢的時候直接將兩個異或和異或起來就好了

關於子樹上的點到根節點路徑重複的問題？

因爲我們可以知道兩個點從他們的lca到根節點的路徑都是重複的

然後兩個值異或起來之後就消掉了，所以我們可以直接將兩個點到根節點的異或值直接異或起來.

code:

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

#define N 100010
#define M 1010

using namespace std;
struct node {
	int next, to, dis;
}edge[N << 1];
struct tree {
	int x, y, dis;
}bian[N];
int n, m, q, head[N << 1], add_edge;
int fath[N], dis[N];

int read() {
	int s = 0, f = 0; char ch = getchar();
	while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
	while (isdigit(ch)) s = s * 10 + (ch ^ 48), ch = getchar();
	return f ? -s : s;
}

void add(int from, int to, int dis) {
	edge[++add_edge].next = head[from];
	edge[add_edge].to = to;
	edge[add_edge].dis = dis;
	head[from] = add_edge;
}

bool cmp(tree a, tree b) {
	return a.dis < b.dis;
}

int father(int x) {
	if (x != fath[x]) fath[x] = father(fath[x]);
	return fath[x];
}

void kruskal() {
	for (int i = 1; i <= n; i++) fath[i] = i;
	int cnt = 0;
	for (int i = 1; i <= m; i++) {
		if (father(bian[i].x) != father(bian[i].y)) {
			fath[bian[i].x] = bian[i].y;
			add(bian[i].x, bian[i].y, bian[i].dis);
			add(bian[i].y, bian[i].x, bian[i].dis);
			cnt ++;
		}
		if (cnt == n - 1) break;
	}
}

void dfs(int x, int fa) {
	for (int i = head[x]; i; i = edge[i].next) {
		int to = edge[i].to;
		if (to == fa) continue;
		dis[to] = (dis[x] ^ edge[i].dis);
		dfs(to, x);
	}
}

int main() {
	n = read(), m = read(), q = read();
	for (int i = 1; i <= m; i++) 
		bian[i].x = read(), bian[i].y = read(), bian[i].dis = read();
	kruskal();
	dfs(1, 0);
	for (int i = 1, x, y; i <= q; i++) {
		x = read(), y = read();
		printf("%d\n", (dis[x] ^ dis[y]));
	}
	return 0;
}