bzoj4811 [Ynoi2017]由乃的OJ 树链剖分+贪心+二进制

题目传送门

https://lydsy.com/JudgeOnline/problem.php?id=4811

题解

我现在为什么都写一题，调一天啊，马上真的退役不花一分钱了。

分成每一位来考虑，如果高位可以是 \(1\) 的话，那么尽量让高位为 \(1\)。

求出当前位为 \(0/1\) 时，最终得到的是 \(0\) 还是 \(1\)。因为要保证选的数小于 \(z\)，所以对于都可以得到 \(1\) 的情况，尽量选择 \(0\) 可以解除限制。

如果 \(z\) 这一位为 \(0\) 并且现在仍然被限制着，那么显然只能取 \(0\) 了。

但是这样做是 \(O(mk\log^2n)\) 的，无法通过 dllxl 的数据。

考虑如何优化。

也就是 \(s_0\) 表示初值为 \(0\) 的结果，\(s_1\) 表示初值为 \(111111..111\) 的结果。

这样做就可以去掉一个 \(k\) 了，时间复杂度 \(O(m(k+\log^2n))\)。

#include<bits/stdc++.h>

#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back

template<typename A, typename B> inline char smax(A &a, const B &b) {return a < b ? a = b , 1 : 0;}
template<typename A, typename B> inline char smin(A &a, const B &b) {return b < a ? a = b , 1 : 0;}

typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;

template<typename I>
inline void read(I &x) {
    int f = 0, c;
    while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
    x = c & 15;
    while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
    f ? x = -x : 0;
}

#define lc o << 1
#define rc o << 1 | 1

const int N = 100000 + 7;

int n, m, k, dfc;
ull S;
int opt[N];
ull v[N];
int dep[N], f[N], siz[N], son[N], dfn[N], pre[N], top[N];

struct Edge { int to, ne; } g[N << 1]; int head[N], tot;
inline void addedge(int x, int y) { g[++tot].to = y, g[tot].ne = head[x], head[x] = tot; }
inline void adde(int x, int y) { addedge(x, y), addedge(y, x); }

struct Node {
    ull s[2], r[2];
    inline Node() : s{0, S}, r{0, S} {}
    inline Node(const int &i) {
        int opt = ::opt[i];
        ull v = ::v[i];
        if (opt == 1) s[0] = 0, s[1] = v;
        else if (opt == 2) s[0] = v, s[1] = S;
        else s[0] = v, s[1] = (~v) & S;
        r[0] = s[0], r[1] = s[1];
    }
    inline Node(const ull &x, const ull &y) : s{x, y}, r{x, y} {}
} t[N << 2];
inline Node operator + (const Node &a, const Node &b) {
    Node ans;
    ans.s[0] = (a.s[0] & b.s[1]) | ((~a.s[0]) & b.s[0]);
    ans.s[1] = (a.s[1] & b.s[1]) | ((~a.s[1]) & b.s[0]);
    ans.r[0] = (b.r[0] & a.r[1]) | ((~b.r[0]) & a.r[0]);
    ans.r[1] = (b.r[1] & a.r[1]) | ((~b.r[1]) & a.r[0]);
    // dbg("******* %llu, %llu;    %llu, %llu;     %llu, %llu\n", a.s[0], a.s[1], b.s[0], b.s[1], ans.s[0], ans.s[1]);
    return ans;
}
inline Node operator - (const Node &a) {
    Node ans;
    ans.s[0] = a.r[0], ans.s[1] = a.r[1];
    ans.r[0] = a.s[0], ans.r[1] = a.s[1];
    return ans;
}

inline void build(int o, int L, int R) {
    if (L == R) return t[o] = Node(pre[L]), (void)0;
    int M = (L + R) >> 1;
    build(lc, L, M), build(rc, M + 1, R);
    t[o] = t[lc] + t[rc];
}
inline void qadd(int o, int L, int R, int x) {
    if (L == R) return t[o] = Node(pre[L]), (void)0;
    int M = (L + R) >> 1;
    if (x <= M) qadd(lc, L, M, x);
    else qadd(rc, M + 1, R, x);
    t[o] = t[lc] + t[rc];
}
inline Node qsum(int o, int L, int R, int l, int r) {
    if (l <= L && R <= r) return t[o];
    int M = (L + R) >> 1;
    if (r <= M) return qsum(lc, L, M, l, r);
    if (l > M) return qsum(rc, M + 1, R, l, r);
    return qsum(lc, L, M, l, r) + qsum(rc, M + 1, R, l, r);
}

inline Node qry(int x, int y) {
    Node ans1, ans2;
    while (top[x] != top[y]) {
        if (dep[top[x]] > dep[top[y]]) {
            ans1 = qsum(1, 1, n, dfn[top[x]], dfn[x]) + ans1;
            x = f[top[x]];
        } else {
            ans2 = qsum(1, 1, n, dfn[top[y]], dfn[y]) + ans2;
            y = f[top[y]];
        }
    }
    // dbg("x = %d, y = %d, dfn: %d %d\n", x, y, dfn[x], dfn[y]);
    if (dep[x] < dep[y]) ans2 = qsum(1, 1, n, dfn[x], dfn[y]) + ans2;
    else ans1 = qsum(1, 1, n, dfn[y], dfn[x]) + ans1;
    ans1 = (-ans1) + ans2;
    return ans1;
}
inline ull solve(int x, int y, ull z) {
    Node a = qry(x, y);
    // dbg("a.s[0] = %llu, a.s[1] = %llu, S = %llu, **** %llu\n", a.s[0], a.s[1], S, 9571068480616515248ull | 16544127868907869972ull);
    // dbg("*** a.s[0] = %llu, a.s[1] = %llu\n", (qry(2, 2)).s[0], (qry(2, 2)).s[1]);
    ull ans = 0, lim = 1;
    for (int i = k - 1; ~i; --i) {
        // dbg("%llu, %llu %llu\n", (z >> i) & 1, ((a.s[0] >> i) & 1), ((a.s[1] >> i) & 1));
        if ((!lim || ((z >> i) & 1)) && !((a.s[0] >> i) & 1) && ((a.s[1] >> i) & 1)) ans |= 1ull << i;
        else ans |= (1ull << i) & a.s[0], lim = lim && !((z >> i) & 1);//, dbg("i = %d\n", i);
    }
    // ull ans2 = 0;
    // for (int i = 0; i <= z; ++i) {
    //     ull cnt = 0;
    //     for (int j = k - 1; ~j; --j) cnt += a.s[(i >> j) & 1] & (1ull << j);
    //     smax(ans2, cnt);
    // }
    return ans;
}

inline void dfs1(int x, int fa = 0) {
    dep[x] = dep[fa] + 1, f[x] = fa, siz[x] = 1;
    for fec(i, x, y) if (y != fa) dfs1(y, x), siz[x] += siz[y], siz[y] > siz[son[x]] && (son[x] = y);
}
inline void dfs2(int x, int pa) {
    top[x] = pa, dfn[x] = ++dfc, pre[dfc] = x;
    if (!son[x]) return; dfs2(son[x], pa);
    for fec(i, x, y) if (y != f[x] && y != son[x]) dfs2(y, y);
}

inline void work() {
    dfs1(1), dfs2(1, 1), build(1, 1, n);
    while (m--) {
        int opt, x, y;
        ull z;
        read(opt), read(x), read(y), read(z);
        if (opt == 2) ::opt[x] = y, v[x] = z, qadd(1, 1, n, dfn[x]);
        else printf("%llu\n", solve(x, y, z));
    }
}

inline void init() {
    read(n), read(m), read(k);
    if (k < 64) S = (1ull << k) - 1;
    else S = -1;
    // dbg("k = %d, S = %llu, %llu\n", k, (1ull << (k - 0)), 1ull << 64);
    for (int i = 1; i <= n; ++i) read(opt[i]), read(v[i]);
    int x, y;
    for (int i = 1; i < n; ++i) read(x), read(y), adde(x, y);
}

int main() {
#ifdef hzhkk
    freopen("hkk.in", "r", stdin);
#endif
    init();
    work();
    fclose(stdin), fclose(stdout);
    return 0;
}

原文地址：https://www.cnblogs.com/hankeke/p/bzoj4811.html