//parent为并查集,FIND为并查集的查找操作
//QUERY为询问结点对集合
//TREE为基图有根树Tarjan(u)
visit[u] = true
for each (u, v) in QUERY
if visit[v]
ans(u, v) = FIND(v)
for each (u, v) in TREE
if !visit[v]
Tarjan(v)
parent[v]=u——————————————————————————————————————————————————————————————————————————————-
#include <iostream>
#include <string.h>
#include <stdio.h>
#define NN 40002 // number of house
#define MM 202 // number of query
using namespace std;
typedef struct node{
int v;
int d;
struct node *nxt;
}NODE;
NODE *Link1[NN];
NODE edg1[NN * 2]; // 树中的边
NODE *Link2[NN];
NODE edg2[NN * 2]; // 询问的点对
int idx1, idx2, N, M;
int res[MM][3]; // 记录结果,res[i][0]: u res[i][1]: v res[i][2]: lca(u, v)
int fat[NN];
int vis[NN];
int dis[NN];
void Add(int u, int v, int d, NODE edg[], NODE *Link[], int &idx){
edg[idx].v = v;
edg[idx].d = d;
edg[idx].nxt = Link[u];
Link[u] = edg + idx++;
edg[idx].v = u;
edg[idx].d = d;
edg[idx].nxt = Link[v];
Link[v] = edg + idx++;
}
int find(int x){ // 并查集路径压缩
if(x != fat[x]){
return fat[x] = find(fat[x]);
}
return x;
}
void Tarjan(int u){
vis[u] = 1;
fat[u] = u;
for (NODE *p = Link2[u]; p; p = p->nxt){
if(vis[p->v]){
res[p->d][2] = find(p->v); // 存的是最近公共祖先结点
}
}
for (NODE *p = Link1[u]; p; p = p->nxt){
if(!vis[p->v]){
dis[p->v] = dis[u] + p->d;
Tarjan(p->v);
fat[p->v] = u;
}
}
}
int main() {
int T, i, u, v, d;
scanf("%d", &T);
while(T--){
scanf("%d%d", &N, &M);
idx1 = 0;
memset(Link1, 0, sizeof(Link1));
for (i = 1; i < N; i++){
scanf("%d%d%d", &u, &v, &d);
Add(u, v, d, edg1, Link1, idx1);
}
idx2 = 0;
memset(Link2, 0, sizeof(Link2));
for (i = 1; i <= M; i++){
scanf("%d%d", &u, &v);
Add(u, v, i, edg2, Link2, idx2);
res[i][0] = u;
res[i][1] = v;
}
memset(vis, 0, sizeof(vis));
dis[1] = 0;
Tarjan(1);
for (i = 1; i <= M; i++){
printf("%d\n", dis[res[i][0]] + dis[res[i][1]] - 2 * dis[res[i][2]]);
}
}
return 0;
}