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.
代码：
#include
#include
#include
#include
#include
#include
#define N 10001
using namespace std;
vector
int n,m,x,y,z,fa[N],deep[N],dis[N],size[N],top[N];
int lca(int x,int y)
{
while(top[x]!=top[y])
{
if(deep[x]
x=fa[top[x]];
}
if(deep[x]>deep[y]) swap(x,y);
return x;
}
int dfs(int x)
{
size[x]=1;
deep[x]=deep[fa[x]]+1;
for(int i=0;i
if(fa[x]!=vec[x][i].first)
{
fa[vec[x][i].first]=x;
dis[vec[x][i].first]=dis[x]+vec[x][i].second;
dfs(vec[x][i].first);
size[x]+=size[vec[x][i].first];
}
}
}
int dfs1(int x)
{
int t=0;
if(!top[x]) top[x]=x;
for(int i=0;i
if(t) top[t]=top[x],dfs1(t);
for(int i=0;i
dfs1(vec[x][i].first);
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i
scanf("%d%d%d",&x,&y,&z);
vec[x].push_back(make_pair(y,z));
vec[y].push_back(make_pair(x,z));
}
dfs(1); dfs1(1);
for(int i=1;i<=m;i++)
{
scanf("%d%d",&x,&y);
printf("%d\n",dis[x]+dis[y]-2*dis[lca(x,y)]);
}
return 0;
}