考虑一个朴素的树形 DP：
设 \(f_{u,0/1}\) 表示选 / 不选 \(u\) 这个点的情况下，以 \(u\) 为根子树内的最小点覆盖。
有转移 \[
\begin{align*}
f_{u,0}=&\sum\limits_{(u,v)\in E} f_{v,1} \\
f_{u,1}=&a_u+\sum\limits_{(u,v)\in E} \min(f_{v,0},f_{v,1})
\end{align*}
\]
对轻重儿子分类讨论，设 \(g_{u,0/1}\) 表示不考虑重儿子贡献的 \(f_u\)。
则 \[
\begin{align*}
f_{u,0}=&g_{u,0}+f_{\text{son}_u,1} \\
f_{u,1}=&g_{u,1}+\min(f_{\text{son}_u,0},f_{\text{son}_u,1})
\end{align*}
\]
定义 \(\otimes\) 为把乘法换成加法，把加法换成去 \(\min\) 的矩乘。
把转移写成 \(\otimes\) 的形式： \[
\begin{bmatrix}
\infty&g_{u,0}\\
g_{u,1}&g_{u,1}
\end{bmatrix}
\otimes
\begin{bmatrix}
f_{\text{son}_u,0}\\
f_{\text{son}_u,1}
\end{bmatrix}=
\begin{bmatrix}
f_{u,0}\\
f_{u,1}
\end{bmatrix}
\]
树剖维护即可。
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using namespace std;
const int N = 1e5;
int n,m;
int a[N + 5];
int to[N * 2 + 5],pre[N * 2 + 5],first[N + 5];
inline void add(int u,int v)
{
static int tot = 0;
to[++tot] = v,pre[tot] = first[u],first[u] = tot;
}
struct Matrix
{
long long a[2][2];
inline Matrix()
{
a[0][0] = a[0][1] = a[1][0] = a[1][1] = 0x3f3f3f3f3f3f3f3f;
}
inline Matrix(int)
{
a[0][0] = a[1][1] = 0,a[0][1] = a[1][0] = 0x3f3f3f3f3f3f3f3f;
}
inline Matrix(long long _0,long long _1)
{
a[0][0] = 0x3f3f3f3f3f3f3f3f,a[0][1] = _0,a[1][0] = a[1][1] = _1;
}
inline Matrix operator*(const Matrix &o)
{
Matrix ret;
for(register int i = 0;i < 2;++i)
for(register int j = 0;j < 2;++j)
for(register int k = 0;k < 2;++k)
ret.a[i][j] = min(ret.a[i][j],a[i][k] + o.a[k][j]);
return ret;
}
} seg[(N << 2) + 10];
void insert(int x,Matrix k,int p,int tl,int tr)
{
if(tl == tr)
{
seg[p] = k;
return ;
}
int mid = tl + tr >> 1;
x <= mid ? insert(x,k,ls,tl,mid) : insert(x,k,rs,mid + 1,tr);
seg[p] = seg[ls] * seg[rs];
}
Matrix query(int l,int r,int p,int tl,int tr)
{
if(l <= tl && tr <= r)
return seg[p];
int mid = tl + tr >> 1;
Matrix ret(1);
l <= mid && (ret = ret * query(l,r,ls,tl,mid),1);
r > mid && (ret = ret * query(l,r,rs,mid + 1,tr),1);
return ret;
}
int fa[N + 5],sz[N + 5],son[N + 5],top[N + 5],id[N + 5],ed[N + 5];
long long f[N + 5][2],g[N + 5][2];
void dfs1(int p)
{
sz[p] = 1,f[p][1] = a[p];
for(register int i = first[p];i;i = pre[i])
if(to[i] ^ fa[p])
{
fa[to[i]] = p,dfs1(to[i]),sz[p] += sz[to[i]];
if(!son[p] || sz[to[i]] > sz[son[p]])
son[p] = to[i];
f[p][0] += f[to[i]][1],f[p][1] += min(f[to[i]][0],f[to[i]][1]);
}
}
void dfs2(int p)
{
static int tot = 0;
id[p] = ++tot,g[p][1] = a[p];
if(son[p])
top[son[p]] = top[p],dfs2(son[p]);
else
ed[top[p]] = id[p];
for(register int i = first[p];i;i = pre[i])
if(!id[to[i]])
top[to[i]] = to[i],g[p][0] += f[to[i]][1],g[p][1] += min(f[to[i]][0],f[to[i]][1]),dfs2(to[i]);
}
void update(int x)
{
for(;x;x = fa[x])
{
insert(id[x],Matrix(g[x][0],g[x][1]),1,1,n),x = top[x];
Matrix cur = query(id[x],ed[x],1,1,n);
g[fa[x]][0] -= f[x][1],g[fa[x]][1] -= min(f[x][0],f[x][1]);
f[x][0] = cur.a[0][1],f[x][1] = cur.a[1][1];
g[fa[x]][0] += f[x][1],g[fa[x]][1] += min(f[x][0],f[x][1]);
}
}
int main()
{
freopen("defense.in","r",stdin),freopen("defense.out","w",stdout);
scanf("%d%d%*s",&n,&m);
for(register int i = 1;i <= n;++i)
scanf("%d",a + i);
int u,v;
for(register int i = 1;i < n;++i)
scanf("%d%d",&u,&v),add(u,v),add(v,u);
top[1] = 1,dfs1(1),dfs2(1);
for(register int i = 1;i <= n;++i)
insert(id[i],Matrix(g[i][0],g[i][1]),1,1,n);
int x,y;
for(long long _0,_1;m;--m)
{
scanf("%d%d%d%d",&u,&x,&v,&y),x ^= 1,y ^= 1;
_0 = g[u][x],_1 = g[v][y];
g[u][x] = 0x3f3f3f3f3f3f3f3f,update(u);
g[v][y] = 0x3f3f3f3f3f3f3f3f,update(v);
printf("%lld\n",min(f[1][0],f[1][1]) < 0x3f3f3f3f3f3f3f3f ? min(f[1][0],f[1][1]) : -1LL);
g[u][x] = _0,update(u);
g[v][y] = _1,update(v);
}
}