洛谷 P5651 【基础最短路练习题】

P5651

题目大意

求一个x到y简单路径的异或和.

思路：

因为图中肯定有环我们肯定是不能走环的.

然后我就随便找了一颗生成树，然后在生成树上找到每一个点到根节点
异或和，查询的时候直接将两个异或和异或起来就好了

关于子树上的点到根节点路径重复的问题？

因为我们可以知道两个点从他们的lca到根节点的路径都是重复的

然后两个值异或起来之后就消掉了，所以我们可以直接将两个点到根节点的异或值直接异或起来.

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;
}