洛谷 P2912 [USACO08OCT]牧场散步Pasture Walking

题目描述

The N cows (2 <= N <= 1,000) conveniently numbered 1..N are grazing among the N pastures also conveniently numbered 1..N. Most conveniently of all, cow i is grazing in pasture i.

Some pairs of pastures are connected by one of N-1 bidirectional walkways that the cows can traverse. Walkway i connects pastures A_i and B_i (1 <= A_i <= N; 1 <= B_i <= N) and has a length of L_i (1 <= L_i <= 10,000).

The walkways are set up in such a way that between any two distinct pastures, there is exactly one path of walkways that travels between them. Thus, the walkways form a tree.

The cows are very social and wish to visit each other often. Ever in a hurry, they want you to help them schedule their visits by computing the lengths of the paths between 1 <= L_i <= 10,000 pairs of pastures (each pair given as a query p1,p2 (1 <= p1 <= N; 1 <= p2 <= N).

POINTS: 200

有N(2<=N<=1000)头奶牛，编号为1到W，它们正在同样编号为1到N的牧场上行走.为了方 便，我们假设编号为i的牛恰好在第i号牧场上.

有一些牧场间每两个牧场用一条双向道路相连，道路总共有N - 1条，奶牛可以在这些道路 上行走.第i条道路把第Ai个牧场和第Bi个牧场连了起来(1 <= A_i <= N; 1 <= B_i <= N)，而它的长度 是 1 <= L_i <= 10,000.在任意两个牧场间，有且仅有一条由若干道路组成的路径相连.也就是说，所有的道路构成了一棵树.

奶牛们十分希望经常互相见面.它们十分着急，所以希望你帮助它们计划它们的行程，你只 需要计算出Q(1 < Q < 1000)对点之间的路径长度•每对点以一个询问p1,p2 (1 <= p1 <= N; 1 <= p2 <= N). 的形式给出.

输入输出格式

输入格式：

• Line 1: Two space-separated integers: N and Q

• Lines 2..N: Line i+1 contains three space-separated integers: A_i, B_i, and L_i

• Lines N+1..N+Q: Each line contains two space-separated integers representing two distinct pastures between which the cows wish to travel: p1 and p2

输出格式：

• Lines 1..Q: Line i contains the length of the path between the two pastures in query i.

输入输出样例

输入样例#1：

4 2 
2 1 2 
4 3 2 
1 4 3 
1 2 
3 2 

输出样例#1：

2 
7 

说明

Query 1: The walkway between pastures 1 and 2 has length 2.

Query 2: Travel through the walkway between pastures 3 and 4, then the one between 4 and 1, and finally the one between 1 and 2, for a total length of 7.

LCA裸题 

屠龙宝刀点击就送

#include <ctype.h>
#include <cstdio>
#define N 1005 

void read(int &x)
{
    x=0;register char ch=getchar();
    for(;!isdigit(ch);ch=getchar());
    for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
}
struct Edge
{
    int next,to,dis;
    Edge (int next=0,int to=0,int dis=0) :next(next),to(to),dis(dis){}
}edge[N<<1];
int dis[N],dad[N][25],dep[N],head[N],cnt,n,q;
void insert(int u,int v,int w)
{
    edge[++cnt]=Edge(head[u],v,w);
    head[u]=cnt;
}
void swap(int &x,int &y)
{
    int tmp=y;
    y=x;
    x=tmp;
}
void dfs(int x)
{
    dep[x]=dep[dad[x][0]]+1;
    for(int i=0;dad[x][i];i++)
    dad[x][i+1]=dad[dad[x][i]][i];
    for(int u=head[x];u;u=edge[u].next)
    {
        int v=edge[u].to;
        if(dad[x][0]!=v)
        {
            dad[v][0]=x;
            dis[v]=dis[x]+edge[u].dis;
            dfs(v);
        }
    }
}
int lca(int x,int y)
{
    if(dep[x]>dep[y]) swap(x,y);
    for(int i=20;i>=0;i--)
    if(dep[dad[y][i]]>=dep[x]) y=dad[y][i];
    if(x==y) return x;
    for(int i=20;i>=0;i--)
    if(dad[y][i]!=dad[x][i]) y=dad[y][i],x=dad[x][i];
    return dad[x][0];
}
int main()
{
    read(n);
    read(q);
    for(int x,y,z,i=1;i<n;i++)
    {
        read(x);
        read(y);
        read(z);
        insert(x,y,z);
        insert(y,x,z);
    }
    dfs(1);
    for(int x,y;q--;)
    {
        read(x);
        read(y);
        int LCA=lca(x,y);
        printf("%d
",dis[x]+dis[y]-2*dis[LCA]);
    }
    return 0;
}