在一棵树上 求2个点的最短距离。那么首先利用LCA找到2个点的近期公共祖先

公式：ans = dis(x) + dis(y) - 2 * dis(lca(x,y))

这里的dis(x)指的上x距离根节点的距离

注意一些细节方面，比方数组的越界问题：

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn = 45555;
struct Edge{
    int to;
    LL  dist;
    Edge(int to,LL dist):to(to),dist(dist){};
};
int n,m;
int deep[maxn],pa[maxn][22];
LL  dis[maxn];
vector<Edge>G[maxn];
void init(){
    memset(pa,-1,sizeof(pa));
    for(int i = 1; i <= n; i++) G[i].clear();
}
//----------------LAC---------------------------
void dfs(int pos,int d,LL dist){
    //printf("[%d %d]\n",pos,dist);
    deep[pos] = d;
    dis[pos] = dist;
    int Size = G[pos].size();
    for(int i = 0; i < Size; i++)
        dfs(G[pos][i].to,d + 1,dist + G[pos][i].dist);
}
void lca_init(){
    for(int j = 1; (1 << j) <= n; j++)
        for(int i = 1; i <= n; i++)
            if(pa[i][j - 1] != -1)
                pa[i][j] = pa[pa[i][j - 1]][j - 1];
}
int lca(int a,int b){
    if(a == b)
        return a;
    if(deep[a] < deep[b])   swap(a,b);
    int i;
    for(i = 0;(1 << i) <= deep[a]; i++);
    for(int j = i; j >= 0; j--)
        if(pa[a][j] != -1 && deep[pa[a][j]] >= deep[b])
            a = pa[a][j];
    if(a == b)
        return b;
    for(int j = i; j >= 0; j--)
        if(pa[a][j] != -1 && deep[pa[a][j]] != deep[pa[b][j]]){
            a = pa[a][j];
            b = pa[b][j];
        }
    return pa[a][0];
}
//-------------------------------------------------
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        int x,y,z;
        init();
        for(int i = 0; i < n - 1; i++){
            scanf("%d%d%d",&x,&y,&z);
            G[x].push_back(Edge(y,z));
            pa[y][0] = x;
        }
        for(int i = 1; i <= n; i++)if(pa[i][0] == -1){
            dfs(i,0,0);
            break;
        }
        lca_init();
        for(int i = 0; i < m; i++){
            scanf("%d%d",&x,&y);
            LL ans = dis[x] + dis[y] - 2 * dis[lca(x,y)];
            printf("%I64d\n",ans);
        }
    }
    return 0;
}