BZOJ 4712 洪水

首先考虑一个暴力的 DP：
设 fu 表示以 u 为根的子树内，要使得 u 与所有叶子不连通，最小的需要删除的点的点权和。

fu=min(au,∑(u,v)∈Efv)

当 u 没有儿子时右边那个和式当做 ∞ 处理。

按照套路设一个 gu 使得 fu=min(au,fsonu+gu)

定义 ⊗ 为把乘法换成加法，加法换成取 min 的矩阵乘法。
则 [guau∞0]⊗[fsonu0]=[fu0]

树剖维护一下就 OK 了。

代码：

#include <cstdio>
#include <algorithm>
#define ls (p << 1)
#define rs (ls | 1)
using namespace std;
const int N = 2e5;
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 x,long long y)
    {
        a[0][0] = x,a[0][1] = y,a[1][0] = 0x3f3f3f3f3f3f3f3f,a[1][1] = 0;
    }
    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;
    }
};
Matrix 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],g[N + 5];
void dfs1(int p)
{
    sz[p] = 1;
    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] += f[to[i]];
        }
    if(!son[p])
        f[p] = a[p];
    else
        f[p] = min(f[p],(long long)a[p]);
}
void dfs2(int p)
{
    static int tot = 0;
    id[p] = ++tot;
    if(son[p])
        top[son[p]] = top[p],dfs2(son[p]),ed[p] = ed[son[p]];
    else
        ed[p] = id[p],g[p] = 0x3f3f3f3f3f3f3f3f;
    for(register int i = first[p];i;i = pre[i])
        if(!id[to[i]])
            top[to[i]] = to[i],g[p] += f[to[i]],dfs2(to[i]);
}
void update(int x)
{
    for(;x;x = fa[x])
    {
        insert(id[x],Matrix(g[x],a[x]),1,1,n),x = top[x];
        Matrix cur = query(id[x],ed[x],1,1,n);
        g[fa[x]] -= f[x];
        f[x] = min(cur.a[0][0],cur.a[0][1]);
        g[fa[x]] += f[x];
    }
}
int main()
{
    scanf("%d",&n);
    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],a[i]),1,1,n);
    scanf("%d",&m);
    char op;
    for(;m;--m)
    {
        scanf(" %c%d",&op,&u);
        if(op == 'Q')
        {
            Matrix ans = query(id[u],ed[u],1,1,n);
            printf("%lld\n",min(ans.a[0][0],ans.a[0][1]));
        }
        else
            scanf("%d",&v),a[u] += v,update(u);
    }
}