HDU 2874 Connections between cities(LCA Tarjan)

Connections between cities

【题目链接】Connections between cities

【题目类型】LCA Tarjan

&题意：

输入一个森林，总节点不超过N(N<10000),由C次询问（C<1000000）,每次询问两个点，如果来联通输出，两点之间的距离，如果不来联通，输出“Not connected”

&题解：

md,就没人吐槽这题询问时有相等的情况吗?我被这个坑了一天= = 最后把ans数组置为-1才过的,以前一直初始化为0,就总是wa,因为有ans0的情况,就是在ij时.

【时间复杂度】\(O(n+q)\)

&代码：

#include <cstdio>
#include <bitset>
#include <iostream>
#include <set>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
using namespace std;
#define INF 0x3f3f3f3f
#define ll long long
#define fo(i,a,b) for(int i=(a);i<=(b);i++)
#define fd(i,a,b) for(int i=(a);i>=(b);i--)
#define cle(a,v) memset(a,(v),sizeof(a))
const int maxn = 2e4 + 7, maxq = 2e6 + 7;
int n, m, q, Head[maxn], QHead[maxn], Dis[maxn], id, iq, fa[maxn], ans[maxq / 2], vis[maxn];
int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
struct Edge {
	int to, next, w;
} ed[maxn], Qed[maxq];
void ori() {
	cle(Head, -1); cle(QHead, -1);
	id = iq = 0;
	cle(vis, 0);
	cle(Dis, 0);
	cle(ans, -1);
}
void Add(int u, int v, int w) {
	ed[id].to = v; ed[id].w = w; ed[id].next = Head[u];
	Head[u] = id++;
}
void QAdd(int u, int v, int index) {
	Qed[iq].to = v, Qed[iq].next = QHead[u], Qed[iq].w = index;
	QHead[u] = iq++;
}
void LCA(int r, int d, int top) {
	fa[r] = r;
	vis[r] = top;
	Dis[r] = d;
	for(int k = Head[r]; ~k; k = ed[k].next) {
		if(!vis[ed[k].to]) {
			LCA(ed[k].to, d + ed[k].w, top);
			fa[ed[k].to] = r;
		}
	}
	for(int k = QHead[r]; ~k; k = Qed[k].next) {
		if(vis[Qed[k].to] == top) {
			ans[Qed[k].w] = Dis[r] + Dis[Qed[k].to] - 2 * Dis[find(Qed[k].to)];
		}
	}
}
int main() {
	freopen("E:1.in", "r", stdin);
	while(~scanf("%d%d%d", &n, &m, &q)) {
		ori();
		int u, v, w;
		fo(i, 1, m) {
			scanf("%d%d%d", &u, &v, &w);
			Add(u, v, w); Add(v, u, w);
		}
		fo(i, 1, q) {
			scanf("%d%d", &u, &v);
			QAdd(u, v, i); QAdd(v, u, i);
		}
		fo(i, 1, n) {
			if(!vis[i]) {
				LCA(i, 0, i);
			}
		}
		fo(i, 1, q) {
			if(ans[i] == -1) printf("Not connected\n");
			else printf("%d\n", ans[i]);
		}
	}
	return 0;
}