poj 1986 Distance Queries 带权lca 模版题

Distance Queries

 

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

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. 

模版题

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
#define true ture
#define false flase
using namespace std;
#define ll long long
#define inf 0xfffffff
int scan()
{
    int res = 0 , ch ;
    while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
    {
        if( ch == EOF )  return 1 << 30 ;
    }
    res = ch - '0' ;
    while( ( ch = getchar() ) >= '0' && ch <= '9' )
        res = res * 10 + ( ch - '0' ) ;
    return res ;
}
#define maxn 100010
#define M 22
struct is
{
    int v,next,w;
} edge[maxn*2];
int deep[maxn],jiedge;
int dis[maxn];
int head[maxn];
int rudu[maxn];
int fa[maxn][M];
void add(int u,int v,int w)
{
    jiedge++;
    edge[jiedge].v=v;
    edge[jiedge].w=w;
    edge[jiedge].next=head[u];
    head[u]=jiedge;
}
void dfs(int u)
{
    for(int i=head[u]; i; i=edge[i].next)
    {
        int v=edge[i].v;
        int w=edge[i].w;
        if(!deep[v])
        {
            dis[v]=dis[u]+edge[i].w;
            deep[v]=deep[u]+1;
            fa[v][0]=u;
            dfs(v);
        }
    }
}
void st(int n)
{
    for(int j=1; j<M; j++)
        for(int i=1; i<=n; i++)
            fa[i][j]=fa[fa[i][j-1]][j-1];
}
int LCA(int u , int v)
{
    if(deep[u] < deep[v]) swap(u , v) ;
    int d = deep[u] - deep[v] ;
    int i ;
    for(i = 0 ; i < M ; i ++)
    {
        if( (1 << i) & d )  // 注意此处，动手模拟一下，就会明白的
        {
            u = fa[u][i] ;
        }
    }
    if(u == v) return u ;
    for(i = M - 1 ; i >= 0 ; i --)
    {
        if(fa[u][i] != fa[v][i])
        {
            u = fa[u][i] ;
            v = fa[v][i] ;
        }
    }
    u = fa[u][0] ;
    return u ;
}
void init()
{
    memset(head,0,sizeof(head));
    memset(fa,0,sizeof(fa));
    memset(rudu,0,sizeof(rudu));
    memset(deep,0,sizeof(deep));
    jiedge=0;
}
int main()
{
    int x,n,t;
    while(~scanf("%d%d",&n,&x))
    {
        init();

        for(int i=0; i<x; i++)
        {
            char a[2];
            int u,v,w;
            scanf("%d%d%d %s",&u,&v,&w,a);
            add(u,v,w);
            add(v,u,w);//双向可以从任意点开始，并且避免有环
        }
                deep[1]=1;
                dis[1]=0;
                dfs(1);
        st(n);
        scanf("%d",&t);
        while(t--)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            printf("%d
",dis[a]-2*dis[LCA(a,b)]+dis[b]);
        }
    }
    return 0;
}