HDU 2874 Connections between cities(LCA离线)



题意：一个森林，询问两个节点距离，若无法到达，输出Not connected。

思路：还是求LCA的思想，只需再对每个询问的两个节点判断是否在一棵树内即可。

有一个问题是这道题的query很大，达到了1000000，所以离线算法空间上比较虚，

然而只会离线的.....于是把int改成short int险过....

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#define eps 1e-6
#define LL long long
using namespace std;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
const int maxn = 10005;
//const int INF = 0x3f3f3f3f;
int dist[maxn], pnt[maxn], ans[1000005], forest[maxn];
bool vis[maxn];
vector<int> G[maxn], w[maxn], num[maxn];
vector<short int> query[maxn];
int n, m, c;
int find(int x) {
	if(x == pnt[x]) return x;
	return pnt[x] = find(pnt[x]);
}
int find2(int x) {
	if(x == forest[x]) return x;
	return forest[x] = find2(forest[x]);
}
void dfs(int u, int val) {
	dist[u] = val; vis[u] = 1; pnt[u] = u;
	int sz1 = G[u].size();
	for(int i = 0; i < sz1; i++) {
		int v = G[u][i];
		if(vis[v]) continue;
		dfs(v, val+w[u][i]);
		pnt[v] = u;
	}
	int sz2 = query[u].size();
	for(int i = 0; i < sz2; i++) {
		int v = query[u][i];
		if(vis[v]) ans[num[u][i]] = dist[u] + dist[v] - 2*dist[find(v)];
	}
}
void init() {
	memset(vis, 0, sizeof(vis));
	for(int i = 1; i <= n; i++) {
		G[i].clear();
		w[i].clear();
		query[i].clear();
		num[i].clear();
		forest[i] = i;
	}
}
int main() {
//	freopen("input.txt", "r", stdin);
	while(scanf("%d%d%d", &n, &m, &c) == 3) {
		init();
		for(int i = 0; i < m; i++) {
			int u, v, d;
			scanf("%d%d%d", &u, &v, &d);
//			cout << find2(u) << " " << find2(v) << endl;
			forest[find2(u)] = find2(v);
//			cout << find2(u) << " " << find2(v) << endl;
			G[u].push_back(v);
			G[v].push_back(u);
			w[u].push_back(d);
			w[v].push_back(d);
		}
		for(int i = 0; i < c; i++) {
			int u, v;
			scanf("%d%d", &u, &v);
			if(find2(u) != find2(v)) {
			//	cout << find2(u) << " " << find2(v) << endl;
				ans[i] = -1; continue;
			}
			query[u].push_back(v);
			query[v].push_back(u);
			num[u].push_back(i);
			num[v].push_back(i);
		}
		for(int i = 1; i <= n; i++) if(!vis[i]) dfs(i, 0);
	//	for(int i = 0; i < c; i++) cout << ans[i] << endl;
		for(int i = 0; i < c; i++)
			if(ans[i] == -1) printf("Not connected\n");
			else printf("%d\n", ans[i]);
	}
	return 0;
}

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