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

4 2

2 1 2

4 3 2

1 4 3

1 2

3 2

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.

路径是一条树两点间的距离用lca求解

考虑边权问题，由于倍增法从节点1开始dfs处理深度与边权，可以处理出前缀和，那么两点间路径边权和就是dis[a]+dis[b]-2*dis[lca(a,b)]

#include<cstdio>

#include<algorithm>

using namespace std;

const int maxn = ;

int n,m,num,head[maxn];

struct node{

    int v,w,next;

}edge[maxn];int dad[maxn][],deep[maxn],dis[maxn];

inline void add_edge(int u,int v,int w) {

    edge[++num].v=v,edge[num].w=w,edge[num].next=head[u],head[u]=num;

}

void dfs(int x) {

    deep[x]=deep[dad[x][]]+;

    for(int i=;dad[x][i];i++)

        dad[x][i+]=dad[dad[x][i]][i];

    for(int i=head[x];i;i=edge[i].next)

        if(!deep[edge[i].v]) {

            dis[edge[i].v]=dis[x]+edge[i].w;

            //printf("%d - > %d  :  %d\n",x,edge[i].v,edge[i].v);

            dad[edge[i].v][]=x;

            dfs(edge[i].v);

        }

}

int lca(int x,int y) {

    if(deep[x]>deep[y])swap(x,y);

    for(int i=;i>=;i--)

        if(deep[dad[y][i]]>=deep[x])y=dad[y][i];

    if(x==y)return x;

    for(int i=;i>=;i--)

        if(dad[x][i]!=dad[y][i]) {

            x=dad[x][i];

            y=dad[y][i];

        }

    return dad[x][];

}

int main () {

    scanf("%d%d",&n,&m);

    for(int a,b,c,i=;i<n;++i) {

        scanf("%d%d%d",&a,&b,&c);

        add_edge(a,b,c);add_edge(b,a,c);

    }

    dfs();

    int a,b;

    while(m--) {

        scanf("%d%d",&a,&b);

        printf("%d\n",dis[a]+dis[b]-*dis[lca(a,b)]);

        //printf("%d\n",dis[4]);

    }

    return ;

}