luogu 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.

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 = 100010;
int n,m,num,head[maxn];
struct node{
    int v,w,next;
}edge[maxn];int dad[maxn][22],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][0]]+1;
    for(int i=0;dad[x][i];i++) 
        dad[x][i+1]=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
",x,edge[i].v,edge[i].v);
            dad[edge[i].v][0]=x;
            dfs(edge[i].v);
        }
}
int lca(int x,int y) {
    if(deep[x]>deep[y])swap(x,y);
    for(int i=20;i>=0;i--) 
        if(deep[dad[y][i]]>=deep[x])y=dad[y][i];
    if(x==y)return x; 
    for(int i=20;i>=0;i--)
        if(dad[x][i]!=dad[y][i]) {
            x=dad[x][i];
            y=dad[y][i];
        }
    return dad[x][0];
}
int main () {
    scanf("%d%d",&n,&m);
    for(int a,b,c,i=1;i<n;++i) {
        scanf("%d%d%d",&a,&b,&c);
        add_edge(a,b,c);add_edge(b,a,c);
    }
    dfs(1);
    int a,b;
    while(m--) {
        scanf("%d%d",&a,&b);
        printf("%d
",dis[a]+dis[b]-2*dis[lca(a,b)]);
        //printf("%d
",dis[4]);
    }
    return 0;
}