递减类贡献问题，考虑兔队线段树：0118A

http://47.92.197.167:5283/contest/454/problem/1

我们其实就是要在树上从根到某个点的路径维护从任意地方开始的：

而这东西，是典型的兔队线段树形式

我们只需要把询问离线，然后拿线段树维护当前点到根的路径即可

// ubsan: undefined
// accoders
#include
using namespace std;
#ifdef LOCAL
 #define debug(...) fprintf(stdout, ##__VA_ARGS__)
#else
 #define debug(...) void(0)
#endif
#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
#define Z(x) (x)*(x)
#define pb push_back
#define fi first
#define se second
//#define M
#define mo (int)(1e9 + 7)
#define N 300010
struct node { int y, z; }; 
struct quey { int x, id; };
vector<node>G[N]; 
vector<quey>E[N]; 
int n, m, i, j, k, T, u, v, w, ans[N];
int dfn[N], dep[N]; 

struct Segment_tree {
#define ls (k << 1) 
#define rs (k << 1 | 1)
#define mid ((l + r) >> 1)
	int mn[N << 2], s[N << 2], ans[N << 2]; 
	pair<int, int> que(int k, int l, int r, int x, int y, int lst) {
		if(l >= x && r <= y) {
			if(l == r) return {s[k] * lst, mn[k]}; 
			if(lst <= mn[k]) return {s[k] * lst, mn[k]}; 
			if(lst <= mn[ls]) {
				auto t = que(rs, mid + 1, r, x, y, lst); 
				int sum = s[ls] * lst + t.fi; 
				return {sum, min(lst, mn[k])}; 
			}
			else {
				auto t = que(ls, l, mid, x, y, lst); 
				int sum = t.fi + ans[k] - ans[ls]; 
				return {sum, min(lst, mn[k])}; 
			}
			return {0, 0}; 
		}
		int sum = 0, mx = 1e18; 
		if(x <= mid) {
			auto t = que(ls, l, mid, x, y, lst); 
			sum += t.fi; mx = t.se; 
		}
		if(y >= mid + 1) {
			auto t = que(rs, mid + 1, r, x, y, min(lst, mx)); 
			sum += t.fi; mx = min(mx, t.se); 
		}
		return {sum, mx}; 
	}
	void push_up(int k, int l, int r) {
		s[k] = s[ls] + s[rs]; 
		mn[k] = min(mn[ls], mn[rs]); 
		ans[k] = ans[ls] + que(rs, mid + 1, r, mid + 1, r, mn[ls]).fi; 
	}
	void set(int k, int l, int r, int x, int Mn, int S) {
		if(l == r) { 
			mn[k] = Mn; s[k] = S; ans[k] = 0; 
			return ; 
		}
		if(x <= mid) set(ls, l, mid, x, Mn, S); 
		else set(rs, mid + 1, r, x, Mn, S); 
		push_up(k, l, r); 
	}
#undef ls
#undef rs
#undef mid
}Seg;

void dfs1(int x, int fa) {
	dep[x] = dep[fa] + 1; 
	for(auto t : G[x]) if(t.y != fa) dfs1(t.y, x); 
}


void dfs2(int x, int fa, int w) {
	dfn[x] = dep[x]; 
//	debug("%lld : %lld\n", x, dep[x]); 
	Seg.set(1, 1, n, dfn[x], x, w); 
	for(auto t : G[x]) if(t.y != fa) dfs2(t.y, x, t.z); 
	for(auto t : E[x]) {
		int y = t.x; 
//		debug("%lld -> %lld\n", y, x); 
		if(dfn[y] <= 0) { ans[t.id] = -1; continue; }
		ans[t.id] = Seg.que(1, 1, n, dfn[y] + 1, dfn[x], y).fi; 
	}
	dfn[x] = -1; 
}

signed main()
{
//	#ifdef LOCAL
	  freopen("tree.in", "r", stdin);
	  freopen("tree.out", "w", stdout);
//	#endif
//	srand(time(NULL));
//	T=read();
//	while(T--) {
//
//	}
	n = read(); m = read(); 
	for(i = 1; i < n; ++i) {
		u = read(); v = read(); w = read(); 
		G[u].pb({v, w}); G[v].pb({u, w}); 
	}
	dfs1(1, 0); 
	for(i = 1; i <= m; ++i) {
		u = read(); v = read(); 
		if(u == v) { ans[i] = 0; continue; }
//		if(dep[u] > dep[v]) swap(u, v); 
		E[v].pb({u, i}); 
	}
	dfs2(1, 0, 0); 
	for(i = 1; i <= m; ++i) printf("%lld\n", ans[i]); 
	return 0;
}