bzoj5210: 最大连通子块和

传送门

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这题一看就是动态dp，先考虑暴力
s表示不选这个点，f表示选这个点
$s[i]=\max (s[t],f[t])$
$f[i]=\max (0,V_x+\sum f[t])$

一开始依照套路写个矩乘（雾
虽然上面的转移我硬是推出来了，但是我推了之后发现$s$并不支持删除一个子树

还是将轻重儿子分开，g表示轻儿子f之和
重链上g的序列代表了什么呢
惊讶的发现$f_i$就是以i为头的最大子段和,$s_i$就是重链上序列g的最大子段和与轻儿子$s_t$ 取max

这显然可以方便的使用线段树维护。
修改也变成了单点修改，g可以直接加减维护，单点s可以开一个可删堆维护

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#include 
#include 
#include 
using namespace std;
 
#define int long long
#define dd c=getchar()
int read() {int s=0,w=1;char c;while (dd,c>'9' || c<'0') if (c=='-') w=-1;while (c>='0' && c<='9') s=s*10+c-'0',dd;return s*w;}
#undef dd
void write(int x) {if (x<0) putchar('-'),x=-x;if (x>=10) write(x/10);putchar(x%10|'0');}
void wln(int x) {write(x);putchar('\n');}void wsp(int x) {write(x);putchar(' ');}
 
const int N = 2e5 + 7;
 
struct edge {
    int t,nxt;
}e[N<<1];
int n,m,cnt;
int V[N],f[N],g[N],s[N];
char c[10];
//=============================================================
int T;
int head[N], siz[N],fa[N],top[N],bot[N],Bs[N],Ms[N],dfn[N],pos[N];
 
void add(int u, int t) {
    e[++cnt].t = t; e[cnt].nxt = head[u]; head[u] = cnt;
}
 
struct heap {
    priority_queue<int> q1,q2;
    void push(int x) {
        q1.push(x);
    }
    void erase(int x) {
        q2.push(x);
    }
    int top() {
        while (!q1.empty() && !q2.empty() && q1.top() == q2.top()) q1.pop(),q2.pop();
        if (q1.empty()) return 0;
        return q1.top();
    }
}Mx[N];
 
void dfs(int x, int F) {
    siz[x] = 1; fa[x] = F;
    for (int i = head[x]; i; i = e[i].nxt) {
        int t = e[i].t;
        if (t == F) continue;
        dfs(t, x);
        siz[x] += siz[t];
        if (siz[t] > siz[Bs[x]]) Bs[x] = t;
    }
}
 
void dfs1(int x, int tp) {
    dfn[++T] = x; pos[x] = T;
    top[x] = tp; bot[x] = x;
    if (Bs[x]) dfs1(Bs[x], tp), bot[x] = bot[Bs[x]] ,s[x] = s[Bs[x]];
    g[x] = V[x];
    for (int i = head[x]; i; i = e[i].nxt) {
        int t = e[i].t;
        if (t == fa[x] || t == Bs[x]) continue;
        dfs1(t, t);
        g[x] += f[t]; Mx[x].push(s[t]);
    }
    f[x] = max(0ll, g[x] + f[Bs[x]]); s[x] = max(s[x], max(f[x], Mx[x].top()));
}
 
//============================================================
 
struct segtree {
    int lc,rc,s,mx;
    segtree() {}
    segtree operator + (segtree a) {
        segtree c;
        c.lc = max(lc, s + a.lc);
        c.rc = max(a.rc, rc + a.s);
        c.s = s + a.s;
        c.mx = max(rc + a.lc, max(mx, a.mx));
        return c;
    }
    #define ls (ts<<1)
    #define rs (ts<<1|1)
    #define lc(x) (tr[x].lc)
    #define rc(x) (tr[x].rc)
    #define s(x) (tr[x].s)
    #define mx(x) (tr[x].mx)
}tr[N<<2];
 
void build(int ts, int l, int r) {
    if (l == r) {
        int x = dfn[l];
        s(ts) = g[x];
        lc(ts) = rc(ts) = max(0ll, g[x]);
        mx(ts) = max(g[x], Mx[x].top());
        return;
    }
    int mid = (l + r) >> 1;
    build(ls, l, mid); build(rs, mid+1, r);
    tr[ts] = tr[ls] + tr[rs];
}
 
void insert(int ts, int l, int r, int x) {
    if (l == r) {
        x = dfn[l];
        s(ts) = g[x];
        lc(ts) = rc(ts) = max(0ll, g[x]);
        mx(ts) = max(g[x], Mx[x].top());
        return;
    }
    int mid = (l + r) >> 1;
    if (x <= mid) insert(ls, l, mid, x);
    else insert(rs, mid+1, r, x);
    tr[ts] = tr[ls] + tr[rs];
}
 
segtree query(int ts, int l, int r, int L, int R) {
    if (l == L && r == R) return tr[ts];
    int mid = (l + r) >> 1;
    if (R <= mid) return query(ls, l, mid, L, R);
    else if (L > mid) return query(rs, mid+1, r, L, R);
    else return query(ls, l, mid, L, mid) + query(rs, mid+1, r, mid+1, R);
}
 
void change(int x, int v) {
    segtree t1, t2;
    bool flg=0;
    while (x) {
        if (flg) Mx[x].erase(t1.mx), Mx[x].push(t2.mx);
        t1 = query(1, 1, n, pos[top[x]], pos[bot[x]]);
        flg = 1;
        g[x] += v; 
        insert(1, 1, n, pos[x]);
        t2 = query(1, 1, n, pos[top[x]], pos[bot[x]]);
        v = t2.lc - f[top[x]]; f[top[x]] = t2.lc;//lc±íʾµÄ¾ÍÊÇf[i] 
        x = fa[top[x]];
    }
}
signed main() {
//  freopen("5210.in", "r", stdin);
//  freopen("5210.out", "w", stdout);
    n = read(); m = read();
    for (int i = 1; i <= n; i++) V[i] = read();
    for (int i = 1, u, v; i < n; i++) {
        u = read(), v = read();
        add(u, v); add(v, u);
    }
    dfs(1,0);
    dfs1(1,1);
    build(1, 1, n);
    for (int i = 1, x, v; i <= m; i++) {
        scanf("%s",c);
        if (c[0] == 'M') {
            x = read(), v = read();
            change(x, v - V[x]);
            V[x] = v;
        }
        else {
            x = read();
            wln(query(1, 1, n, pos[x], pos[bot[x]]).mx);        
        }
    }
}