HDU2874 Connections between cities【LCA】

题意：求两点间的距离，可能不连通

思路：连不连通，用并查集解决。不连通，我直接给每个连通集建了一个树状数组，不用想T了。其实只要建个虚根，把每个连通集的一个点连过去就行了，真是奇妙。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<stdlib.h>
#include<math.h>
#include<vector>
#include<list>
#include<map>
#include<stack>
#include<queue>
#include<algorithm>
#include<numeric>
#include<functional>
using namespace std;

typedef long long ll;
typedef pair<int,int> pii;
#define lson(x) 2*x
#define rson(x) 2*x+1
const int maxn = 20005;
struct Edge
{
	int to,cost,id,next;
}edge[maxn];
int head[maxn],tot;
int rmq_n;                 // rmq_n -> 线段树的点，2 * n - 1 
int vs[maxn*2],depth[maxn*2],id[maxn];
int tree[maxn*4],bit_n;
int root;
int pre[maxn];

struct data
{
	int l,r,num,dep; //num -> 下标，dep -> depth最小值 
}node[maxn*4];

void init(int x)
{
	memset(head,-1,sizeof head);
	tot = 0;
	for(int i = 0; i <= x; i++)
		pre[i] = i;
}

void addedge(int a,int b,int c)
{
	edge[tot].to = b;
	edge[tot].cost = c;
	edge[tot].next = head[a];
	head[a] = tot++;
}

int lowbit(int k)
{
	return k & -k;
}

void add(int cnt,int val)
{
	while(cnt <= bit_n)       // n -> num of point
	{
		tree[cnt] += val;
		cnt += lowbit(cnt);
	}
}

int fid(int cnt) // 1 ~ k sum
{
	int sum = 0;
	while(cnt)
	{
		sum += tree[cnt];
		cnt -= lowbit(cnt);
	}
	return sum;
}

void pushup(int cnt)
{
	node[cnt].dep = min(node[lson(cnt)].dep,node[rson(cnt)].dep);
	if(node[lson(cnt)].dep < node[rson(cnt)].dep)
		node[cnt].num = node[lson(cnt)].num;
	else
		node[cnt].num = node[rson(cnt)].num;
}

void build(int x,int y, int cnt)
{
	node[cnt].l = x;
	node[cnt].r = y;
	if(x == y)
	{
		node[cnt].num = vs[x];
		node[cnt].dep = depth[x];
		return;
	}
	int mid = (x+y) / 2;
	build(x,mid,lson(cnt));
	build(mid+1,y,rson(cnt));
	pushup(cnt);
}

pii query(int x,int y,int cnt)
{
	pii a,b;
	a.second = 0x3f3f3f3f;
	b.second = 0x3f3f3f3f;
	if(x == node[cnt].l && y == node[cnt].r)
	{
		return make_pair(node[cnt].num,node[cnt].dep);
	}
	int fa = 2*cnt;
	if(x <= node[fa].r)
	{
		if(y <= node[fa].r)
			a = query(x,y,fa);
		else
			a = query(x,node[fa].r,fa);
	}
	fa++;
	if(y >= node[fa].l)
	{
		if(x >= node[fa].l)
			b = query(x,y,fa);
		else
			b = query(node[fa].l,y,fa);
	}
	if(a.second < b.second)
		return a;
	else
		return b;
}

void dfs(int v,int p,int d,int &k)
{
	id[v] = k;
	vs[k] = v;
	depth[k++] = d;
	for(int i = head[v]; i != -1; i = edge[i].next)
	{
		Edge &e = edge[i];
		if(e.to != p)
		{
			add(k,e.cost);
			dfs(e.to,v,d+1,k);
			vs[k] = v;
			depth[k++] = d;
			add(k,-e.cost);
		}
	}
}

void yu(int x)
{
	rmq_n = 2*x-1;
	root = 0;
	bit_n = 2 *(x-1); //走过几条边，算边权和 
	for(int i = 0; i <= bit_n; i++)
		tree[i] = 0;	 
	int k = 1;
	dfs(root,-1,0,k);
	build(1,rmq_n,1);
}

int fa(int x)
{
	if(pre[x] == x)
		return x;
	else
		return pre[x] = fa(pre[x]);
}

void he(int x,int y)
{
	int nx = fa(x);
	int ny = fa(y);
	if(nx != ny)
		pre[nx] = ny;
}

int main(void)
{
	int n,m,q;
	while(scanf("%d%d%d",&n,&m,&q)!=EOF)
	{
		init(n);
		while(m--)
		{
			int a,b,c;
			scanf("%d%d%d",&a,&b,&c);
			he(a,b);
			addedge(a,b,c);
			addedge(b,a,c);
		}
		for(int i = 1; i <= n; i++)
		{
			if(fa(i) == i)
			{
				addedge(0,i,0);
				addedge(i,0,0);
			}
		}
		yu(n+1);
		for(int i = 1; i <= q; i++)
		{
			int a,b;
			scanf("%d%d",&a,&b);
			int x = fa(a);
			int y = fa(b);
			if(x != y)
				printf("Not connected\n");
			else
			{
				pii f = query(min(id[a],id[b]),max(id[a],id[b]),1);
				int lca = f.first;
				int ans = fid(id[a]) + fid(id[b]) - 2 * fid(id[lca]);
				printf("%d\n",ans);
			}
		}
	}
	return 0;
}