# 洛谷 P2912 [USACO08OCT]牧场走走Pasture Walking

## 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.

* len[lca(va, vb）]

`````` 1 #include <bits/stdc++.h>
2
3 const int MAXN = 5000 + 10;
4
6 {
7     x = 0;char ch = getchar();char c = ch;
8     while(ch > '9' || ch < '0')c = ch, ch = getchar();
9     while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
10     if(c == '-')x = -x;
11 }
12
13 inline void swap(int &a, int &b)
14 {
15     int tmp = a;
16     a = b;
17     b = tmp;
18 }
19
20 struct Edge{int u,v,w,next;}edge[MAXN >> 1];
22 int n,m,tmp1,tmp2,tmp3;
23 bool b[MAXN];
24
25 inline void insert(int a, int b, int c)
26 {
29 }
30
31 void dfs(int u, int step)
32 {
33     for(int pos = head[u];pos;pos = edge[pos].next)
34     {
35         int v = edge[pos].v;
36         if(!b[v])
37         {
38             b[v] = true;
39             deep[v] = step + 1;
40             len[v] = len[u] + edge[pos].w;
41             p[v] = u;
42             dfs(v, step + 1);
43         }
44     }
45 }
46
47 inline void yuchuli()
48 {
49     b[root] = true;
50     len[root] = 0;
51     deep[root] = 0;
52     dfs(root, 0);
53     for(int i = 1;(1 << i) <= n;i ++)
54     {
55         for(int j = 1;j <= n;j ++)
56         {
57             p[j][i] = p[p[j][i - 1]][i - 1];
58         }
59     }
60 }
61
62 inline int lca(int va, int vb)
63 {
64     if(deep[va] < deep[vb])swap(va, vb);
65     int M = 0;
66     for(;(1 << M) <= n;M ++);
67     M --;
68     for(int i = M;i >= 0;i --)
69         if(deep[va] - (1 << i) >= deep[vb])
70             va = p[va][i];
71     if(va == vb)return va;
72     for(int i = M;i >= 0;i --)
73         if(p[va][i] != p[vb][i])
74         {
75             va = p[va][i];
76             vb = p[vb][i];
77         }
78     return p[va];
79 }
80
81 int main()
82 {
84     for(int i = 1;i < n;i ++)
85     {
87         insert(tmp1, tmp2, tmp3);
88         insert(tmp2, tmp1, tmp3);
89     }
90     root = 1;
91     yuchuli();
92     for(int i = 1;i <= m;i ++)
93     {
95         printf("%d\n", len[tmp1] + len[tmp2] - 2 * len[lca(tmp1, tmp2)]);
96     }
97     return 0;
98 }
``````

## 标题陈诉

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

1条，红牛能够在这几个道路

<= B_i <= N)，而它的长度 是 1 <= L_i <=
10,000.在自便多个牧场间，有且只有一条由若干道路组成的路线相连.也即是说，全部的道路构成了一棵树.

(1 <= p1 <= N; 1 <= p2 <= N). 的花样给出.

## 主题材料陈说

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

1条，红牛能够在那一个道路

<= B_i <= N)，而它的长短 是 1 <= L_i <=
10,000.在自由三个牧场间，有且唯有一条由若干道路组成的不二等秘书技相连.也正是说，全体的征程构成了一棵树.

(1 <= p1 <= N; 1 <= p2 <= N). 的样式给出.

## 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.

## 输入输出样例

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

``````2
7
``````

## 输入输出格式

• 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

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

## 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

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

{
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];
void insert(int u,int v,int w)
{
}
void swap(int &x,int &y)
{
int tmp=y;
y=x;
x=tmp;
}
void dfs(int x)
{
{
int v=edge[u].to;
{
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(x==y) return x;
for(int i=20;i>=0;i--)
}
int main()
{
for(int x,y,z,i=1;i<n;i++)
{
insert(x,y,z);
insert(y,x,z);
}
dfs(1);
for(int x,y;q--;)
{