Distance Queries（LCA+tarjan）

Distance Queries

Farmer John’s cows refused to run in his marathon since he chose a path much too long for their leisurely lifestyle. He therefore wants to find a path of a more reasonable length. The input to this problem consists of the same input as in “Navigation Nightmare”,followed by a line containing a single integer K, followed by K “distance queries”. Each distance query is a line of input containing two integers, giving the numbers of two farms between which FJ is interested in computing distance (measured in the length of the roads along the path between the two farms). Please answer FJ’s distance queries as quickly as possible!
Input

• Lines 1…1+M: Same format as “Navigation Nightmare”

• Line 2+M: A single integer, K. 1 <= K <= 10,000

• Lines 3+M…2+M+K: Each line corresponds to a distance query and contains the indices of two farms.
  Output

• Lines 1…K: For each distance query, output on a single line an integer giving the appropriate distance.
  Sample Input
  7 6
  1 6 13 E
  6 3 9 E
  3 5 7 S
  4 1 3 N
  2 4 20 W
  4 7 2 S
  3
  1 6
  1 4
  2 6
  Sample Output
  13
  3
  36
  Hint
  Farms 2 and 6 are 20+3+13=36 apart.

ac代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN = 80080;
const int MAXQ = 20020;
 
int father[MAXN],Head[MAXN],QHead[MAXN],Dist[MAXN];
 
struct EdgeNode
{
    int to;
    int next;
    int lca;
}Edges[MAXN];
 
EdgeNode QEdges[MAXN];
 
int find(int x)
{
    if(x != father[x])
        father[x] = find(father[x]);
    return father[x];
}
 
bool vis[MAXN];
 
void LCA(int u)
{
    father[u] = u;
    vis[u] = true;//标记该点 
    for(int k = Head[u]; k != -1; k = Edges[k].next)
    {
        if(!vis[Edges[k].to])//若该点没有被标记 
        {
            Dist[Edges[k].to] = Dist[u] + Edges[k].lca;//将其路的长度相加 
            LCA(Edges[k].to);//递归到连接点，即该点的子结点 
            father[Edges[k].to] = u;
        }
    }
 //将数据处理完（存图）之后，进行运算 
    for(int k = QHead[u]; k != -1; k = QEdges[k].next)
    {
        if(vis[QEdges[k].to])
        {
            //QEdges[k].lca = find(QEdges[k].to);
            QEdges[k].lca = Dist[u] + Dist[QEdges[k].to] - 2*Dist[find(QEdges[k].to)];//find是公共祖先。 
            //两条路之间的距离等于两条路分别到根节点的距离和再减去根节点到两点最近公共祖先的距离的二倍 
            QEdges[k^1].lca = QEdges[k].lca;
            printf("%d    %d\n",k,k^1); 
        }
    }
}
 
int main()
{
    int N,M,K,u,v,w,a,b;
    char s;
    while(~scanf("%d%d",&N,&M))
    {
        memset(father,0,sizeof(father));
        memset(Head,-1,sizeof(Head));
        memset(QHead,-1,sizeof(QHead));
        memset(vis,false,sizeof(vis));
        memset(Edges,0,sizeof(Edges));
        memset(QEdges,0,sizeof(QEdges));
        memset(Dist,0,sizeof(Dist));
        int id = 0;
        for(int i = 0; i < M; ++i)
        {
            scanf("%d%d%d %c",&u,&v,&w,&s);
            Edges[id].to = v;
            Edges[id].lca = w;
            Edges[id].next = Head[u];
            Head[u] = id++;
            Edges[id].to = u;
            Edges[id].lca = w;
            Edges[id].next = Head[v];
            Head[v] = id++;
        }//存图，无向图， 
        scanf("%d",&K);
        int iq = 0;
        for(int i = 0; i < K; ++i)
        {
            scanf("%d%d",&a,&b);
            QEdges[iq].to = b;
            QEdges[iq].next = QHead[a];
            QHead[a] = iq++;
            QEdges[iq].to = a;
            QEdges[iq].next = QHead[b];
            QHead[b] = iq++;
        }
        LCA(1);
        for(int i = 0; i < iq; i+=2)
            printf("%d\n",QEdges[i].lca);
    }
 
    return 0;
}