hdu2874—Connections between cities（LCA）

题目大意：给出n个城市，m条路，不存在环，求任意两个城市间的距离，城市间可能不联通

解题思路：ST+并查集， 并查集判断城市间是否联通，因为给出的可能是森林，将每颗树的根节点指向一个root，构成一棵树。

#include   
#include   
#include   
#include   
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;  
const int N = 10010;
const int M = 25;
const int INF = 99999999;

struct Edge{
	int node,len;
	Edge*next;
}m_edge[N*2];
int Ecnt;
Edge*head[N];
int tot,ver[2*N],R[2*N],first[N],dir[N];
bool vis[N];  
int dp[2*N][M];
int father[N];

//ver:节点编号 R：深度 first：点编号位置 dir：距离  
void dfs(int u ,int dep)  
{  
	vis[u] = true; ver[++tot] = u; first[u] = tot; R[tot] = dep;  
	for( Edge*k=head[u]; k ; k = k->next )  
		if( !vis[k->node] ){  
			int v = k->node , w = k->len;  
			dir[v] = dir[u] + w;  
			dfs(v,dep+1);  
			ver[++tot] = u; R[tot] = dep;  
		}  
}  

//对R进行预处理，保存的是R的编号
void ST(int n)  
{  
	for(int i=1;i<=n;i++)  
		dp[i][0] = i;  
	for(int j=1;(1< y) swap(x,y);  
	int res = RMQ(x,y);  
	return ver[res];  
}  

void init()
{
	Ecnt = tot = 0;
	fill( head , head+N , (Edge*)0 );
	fill( vis , vis+N , false );
	for( int i = 0 ; i < N ; ++i )
		father[i] = i;
}

int find( int a )
{
	int t = a;
	while( father[t] != t ){
		father[t] = father[father[t]];
		t = father[t];
	}
	return t;
}

void merge( int a , int b )
{
	int t1 = find( a );
	int t2 = find( b );
	if( t1 != t2 ) father[t1] = t2;
}

void mkEdge( int a , int b , int c )
{
	m_edge[Ecnt].node = a;
	m_edge[Ecnt].len = c;
	m_edge[Ecnt].next = head[b];
	head[b] = m_edge + Ecnt++;

	m_edge[Ecnt].node = b;
	m_edge[Ecnt].len = c;
	m_edge[Ecnt].next = head[a];
	head[a] = m_edge + Ecnt++;
}

int main()
{
	int n,m,p,a,b,c;
	while( ~scanf("%d%d%d",&n,&m,&p) ){
		init();
		for( int i = 0 ; i < m ; ++i ){
			scanf("%d%d%d",&a,&b,&c);
			mkEdge( a , b , c );
			merge( a , b );
		}    
		for( int i = 1 ; i <= n ; ++i ){
			if(father[i]!=i) continue;
			mkEdge( 0 , i , 0 );           //root为0，长度为0
		}
		dir[0] = 0;
		dfs( 0 , 1 );
		ST( 2*(n+1)-1 );    
		for( int i = 0 ; i < p ; ++i ){
			scanf("%d%d",&a,&b);
			if( find(a) != find(b) ){
				printf("Not connected\n"); continue;
			}
			c = LCA(a,b);
			printf("%d\n",dir[a]+dir[b]-2*dir[c]);
		}
	}
	return 0;
}