POJ __1986 Distance Queries （lca+求距离的骚操作）

Distance Queries

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 15728		Accepted: 5536
Case Time Limit: 1000MS

Description

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

Sample Output

13
3
36

Hint

Farms 2 and 6 are 20+3+13=36 apart.

Source

USACO 2004 February

有n个点m个边~a距离b为c（距离root为线性）~后面那个字符没用~询问x个~求dis[i]+dis[j]-2*dis[lca(i,j)];

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int maxn=230100;
const int DEG=20;
struct ***
{
    int to,ne,len;
}ed[maxn*2];
int head[maxn],cnt;
int dis[maxn];//计算距离；
void add(int u,int v,int w)
{
    ed[cnt].to=v;
    ed[cnt].len=w;
    ed[cnt].ne=head[u];
    head[u]=cnt++;
}
void init()
{
    cnt=1;
    memset(head,-1,sizeof(head));
    memset(dis,0,sizeof(dis));
}
int fa[maxn][DEG];
int deg[maxn];//深度数组
void BFS(int root)
{
    queue<int>q;
    deg[root]=0;
    fa[root][0]=root;
    q.push(root);
    dis[root]=0;//
    while(!q.empty())
    {
        int t=q.front();
        q.pop();
        for(int s=1;s<DEG;s++)
        {
            fa[t][s]=fa[fa[t][s-1]][s-1];
        }
        for(int s=head[t];s!=-1;s=ed[s].ne)
        {
            int v=ed[s].to;
            if(v==fa[t][0])
            {
                continue;
            }
            deg[v]=deg[t]+1;
            dis[v]=dis[t]+ed[s].len;//
            fa[v][0]=t;
            q.push(v);
        }
    }
}
int LCA(int u,int v)
{
    if(deg[u]>deg[v])
    {
        swap(u,v);
    }
    int hu=deg[u],hv=deg[v];
    int tu=u,tv=v;
    for(int det=hv-hu,s=0;det;det>>=1,s++)
    {
        if(det&1)
        {
            tv=fa[tv][s];
        }
    }
    if(tu==tv)
    {
        return tu;
    }
    for(int s=DEG-1;s>=0;s--)
    {
        if(fa[tu][s]==fa[tv][s])
        {
            continue;
        }
        tu=fa[tu][s];
        tv=fa[tv][s];
    }
    return fa[tu][0];
}
bool in[maxn];
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    init();
    memset(in,0,sizeof(in));
    for(int s=1;s<=m;s++)
    {
        int a,b,c;
        char w;
        scanf("%d%d%d %c",&a,&b,&c,&w);
        add(a,b,c);
        add(b,a,c);
        in[b]=1;
    }
    int root;
    for(int s=1;s<=n;s++)
    {
        if(!in[s])
        {
            root=s;
            break;
        }
    }
    BFS(root);
    int te;
    cin>>te;
    while(te--)
    {
        int u,v;
        scanf("%d%d",&u,&v);
      //  cout<<LCA(u,v)<<endl;
        cout<<dis[u]+dis[v]-2*dis[LCA(u,v)]<<endl;
    }
    return 0;
}