題面

Bzoj

題解

對於每個節點，我們可以用樹鏈剖分和線段樹維護以下信息：

• 單獨在某個點分配\(i\)個人的最大收益（可以\(O(m)\)計算）
• 分配\(i\)的最大收益（可以\(O(m^2)\)計算）

#include <cstdio>
#include <cstring>
#include <algorithm>
using std::min; using std::max;
using std::sort; using std::swap;
typedef long long ll;

template<typename T>
void read(T &x) {
    int flag = 1; x = 0; char ch = getchar();
    while(ch < '0' || ch > '9') { if(ch == '-') flag = -flag; ch = getchar(); }
    while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); x *= flag;
}

const int N = 20010, M = 51, T = 65600;
int n, m, q, X = 1 << 16, Y = ~0U >> 1, A, B, Q, op, x, y;
int fa[N], son[N], siz[N], top[N], dfn[N], dep[N], val[N], tim;
int cnt, from[N], to[N], nxt[N];
inline void addEdge(int u, int v) {
    to[++cnt] = v, nxt[cnt] = from[u], from[u] = cnt;
}
struct P {
    ll v[M];
    P() { for(int i = 0; i < M; ++i) v[i] = 0; }
    P operator + (P b) {
        P c;
        for(int i = 0; i < M; ++i) c.v[i] = max(v[i], b.v[i]);
        return c;
    }
    P operator * (P b) {
        P c;
        for(int i = 0; i < M; ++i)
            for(int j = 0; j < M - i; ++j)
                c.v[i + j] = max(c.v[i + j], v[i] + b.v[j]);
        return c;
    }
}tmp, a[N], v0[T], v1[T], s0, s1;

//讀入數據
inline int getint() {
    A = ((A ^ B) + B / X + B * X) & Y;
    B = ((A ^ B) + A / X + A * X) & Y;
    return (A ^ B) % Q;
}
inline void gettmp() {
    for(int i = 1; i <= m; ++i) tmp.v[i] = getint();
    sort(&tmp.v[1], &tmp.v[m + 1]);
}

//線段樹
inline void pushup(int o, int lc, int rc) { v0[o] = v0[lc] + v0[rc], v1[o] = v1[lc] * v1[rc]; }
void build(int o = 1, int l = 1, int r = n) {
    if(l == r) { v0[o] = v1[o] = a[val[l]]; return ; }
    int mid = (l + r) >> 1, lc = o << 1, rc = lc | 1;
    build(lc, l, mid), build(rc, mid + 1, r), pushup(o, lc, rc);
}
void modify(int k, int o = 1, int l = 1, int r = n) {
    if(l == r) { v0[o] = v1[o] = tmp; return ; }
    int mid = (l + r) >> 1, lc = o << 1, rc = lc | 1;
    if(k <= mid) modify(k, lc, l, mid);
    else modify(k, rc, mid + 1, r);
    pushup(o, lc, rc);
}
void query0(int ql, int qr, int o = 1, int l = 1, int r = n) {
    if(l >= ql && r <= qr) { s0 = s0 + v0[o]; return ; }
    int mid = (l + r) >> 1, lc = o << 1, rc = lc | 1;
    if(ql <= mid) query0(ql, qr, lc, l, mid);
    if(qr > mid) query0(ql, qr, rc, mid + 1, r);
}
void query1(int ql, int qr, int o = 1, int l = 1, int r = n) {
    if(l >= ql && r <= qr) { s1 = s1 * v1[o]; return ; }
    int mid = (l + r) >> 1, lc = o << 1, rc = lc | 1;
    if(ql <= mid) query1(ql, qr, lc, l, mid);
    if(qr > mid) query1(ql, qr, rc, mid + 1, r);
}

//樹鏈剖分
void dfs(int u) {
    siz[u] = 1, dep[u] = dep[fa[u]] + 1;
    for(int i = from[u]; i; i = nxt[i]) {
        int v = to[i]; dfs(v), siz[u] += siz[v];
        if(siz[v] > siz[son[u]]) son[u] = v;
    }
}
void dfs(int u, int t) {
    dfn[u] = ++tim, val[tim] = u, top[u] = t;
    if(!son[u]) return ; dfs(son[u], t);
    for(int i = from[u]; i; i = nxt[i])
        if(to[i] != son[u]) dfs(to[i], to[i]);
}
inline void Path(int x, int y) { //dep[y] 始終小於等於 dep[x]
    if(x == y) return ;
    x = fa[x]; int fx = top[x];
    while(fx != top[y]) query0(dfn[fx], dfn[x]), x = fa[fx], fx = top[x];
    query0(dfn[y], dfn[x]);
}

int main () {
#ifdef OFFLINE_JUDGE
    freopen("233.in", "r", stdin);
    freopen("233.out", "w", stdout);
#endif
    read(n), read(m), read(A), read(B), read(Q);
    for(int i = 2; i <= n; ++i)
        read(fa[i]), addEdge(fa[i], i);
    for(int i = 1; i <= n; ++i) gettmp(), a[i] = tmp;
    dfs(1), dfs(1, 1), build(), read(q);
    while(q--) {
        read(op), read(x);
        if(!op) gettmp(), modify(dfn[x]);
        else {
            read(y);
            s0 = s1 = P();
            Path(x, y), query1(dfn[x], dfn[x] + siz[x] - 1);
            s0 = s0 * s1;
            printf("%lld\n", s0.v[m]);
        }
    }
    return 0;
}
 

Bzoj2164 采礦（線段樹+樹鏈剖分）