几道有意思的数据结构题
「JOISC 2021 Day3」聚会 2
首先当 \(k\) 为奇数时答案显然为 \(1\)。
否则考虑选择两个子树，把 \(k/2\) 个点放到其中一个子树，剩下的放到另一个中，那么要求就是这两个子树大小必须 \(\ge k/2\)，期待值就是这两颗子树根节点的距离 \(+1\)。
直接点分治，时间复杂度 \(O(n\log^2 n)\)。树状数组在 nkoj 会 TLE，我卡常换成了排序后直接处理，复杂度不变。
好像正解是高贵的 \(O(n\log n)\)，但反正我卡过去了
#pragma GCC optimize("Ofast")
#include <cstdio>
#include <cstring>
#include <algorithm>
#define gc (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 65536, stdin), p1 == p2) ? EOF : *p1 ++)
inline int max(const int x, const int y) {return x > y ? x : y;}
char buf[65536], *p1, *p2;
inline int read() {
char ch;
int x = 0;
while ((ch = gc) < 48);
do x = x * 10 + ch - 48; while ((ch = gc) >= 48);
return x;
}
struct Edge {
int to, nxt;
} e[400005];
int head[200005], sze[200005], sze2[200005], dep[200005], bel[200005], rt, fart, tot, now;
int ans[200005], a[200005], len, n;
bool vis[200005];
inline void AddEdge(int u, int v) {
e[++ tot].to = v, e[tot].nxt = head[u], head[u] = tot;
}
void GetRoot(int u, int fa, int cnt) {
sze2[u] = 1;
int mx = 0;
for (int i = head[u]; i; i = e[i].nxt)
if (e[i].to != fa && !vis[e[i].to])
GetRoot(e[i].to, u, cnt), mx = max(mx, sze2[e[i].to]), sze2[u] += sze2[e[i].to];
mx = max(mx, cnt - sze2[u]);
if (!rt || mx < now) rt = u, fart = fa, now = mx;
}
void getdep(int u, int fa) {
if (u != rt) a[++ len] = u;
sze[u] = 1;
for (int i = head[u]; i; i = e[i].nxt)
if (e[i].to != fa && !vis[e[i].to]) {
dep[e[i].to] = dep[u] + 1;
if (u == rt) bel[e[i].to] = e[i].to;
else bel[e[i].to] = bel[u];
getdep(e[i].to, u);
sze[u] += sze[e[i].to];
}
}
struct cmp {
inline bool operator () (const int x, const int y) const {return sze[x] < sze[y];}
};
void calc(int u) {
std::sort(a + 1, a + len + 1, cmp());
int mx = -1e9, mxb = 0, semx = -1e9, semxb = 0;
for (int i = len; i; -- i) {
if (sze[rt] - sze[bel[a[i]]] >= sze[a[i]])
ans[sze[a[i]] << 1] = max(ans[sze[a[i]] << 1], dep[a[i]]);
if (bel[a[i]] != mxb) ans[sze[a[i]] << 1] = max(ans[sze[a[i]] << 1], dep[a[i]] + mx);
else ans[sze[a[i]] << 1] = max(ans[sze[a[i]] << 1], dep[a[i]] + semx);
if (!mxb || dep[a[i]] > mx) {
if (bel[a[i]] != mxb) semx = mx, semxb = mxb;
mx = dep[a[i]], mxb = bel[a[i]];
}
else if (bel[a[i]] != mxb && (!semxb || dep[a[i]] > semx))
semx = dep[a[i]], semxb = bel[a[i]];
}
}
void dfs(int u, int fa, int cnt) {
rt = len = now = 0, GetRoot(u, -1, cnt), vis[u = rt] = true;
dep[u] = 0, getdep(u, -1), calc(u);
for (int i = head[u]; i; i = e[i].nxt)
if (!vis[e[i].to]) dfs(e[i].to, u, e[i].to == fart ? cnt - sze2[u] : sze2[e[i].to]);
}
int main() {
n = read();
for (int i = 1, u, v; i < n; ++ i)
u = read(), v = read(), AddEdge(u, v), AddEdge(v, u);
dfs(1, -1, n);
for (int i = n - 1; i; -- i) ans[i] = max(ans[i], ans[i + 1]);
for (int i = 1; i <= n; ++ i) printf("%d\n", i & 1 ? 1 : ans[i] + 1);
return 0;
}
「JOISC 2021 Day1」饮食区
把询问离线下来并拆开修改操作，看作在 \(L_i\) 时刻加入该操作，\(R_i\) 时刻删除该操作。把原本操作的时间看成下标。
那么对于当前需要处理的询问 \((a,b,t)\)（\(t\) 是指这是在原始操作序列中的第 \(t\) 个），求出 \([1,t]\) 中的最大后缀和，显然这就是这个询问执行时实际上 \(a\) 队列剩下甜点的个数。
然后线段树上二分出它最早在哪个添加操作后被满足即可。复杂度是优雅的 \(O(n\log n)\)。
#include <cstdio>
#include <vector>
#define gc (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 65536, stdin), p1 == p2) ? EOF : *p1 ++)
inline long long max(const long long x, const long long y) {return x > y ? x : y;}
char buf[65536], *p1, *p2;
inline int read() {
char ch;
int x = 0;
while ((ch = gc) < 48);
do x = x * 10 + ch - 48; while ((ch = gc) >= 48);
return x;
}
inline long long readll() {
char ch;
long long x = 0;
while ((ch = gc) < 48);
do x = x * 10 + ch - 48; while ((ch = gc) >= 48);
return x;
}
struct node {
int l, r;
long long suf, sum, sum2, sum3;
} tree[1000005];
struct tuple {
long long x;
int y, z;
};
struct tuple2 {
long long x, y, z;
};
std::vector<int> vec1[250005], vec2[250005];
std::vector<tuple> vec3[250005];
int ans[250005], idx[250005], c[250005], k[250005], qstot;
void build(int p, int l, int r) {
tree[p].l = l, tree[p].r = r;
if (l != r) {
build(p << 1, l, l + r >> 1);
build(p << 1 | 1, (l + r >> 1) + 1, r);
}
}
void update(int p, int x, int d) {
if (tree[p].l == tree[p].r) {tree[p].suf += d, tree[p].sum += d, tree[p].sum2 = max(0, tree[p].sum3 += d); return;}
int mid = tree[p].l + tree[p].r >> 1;
if (x <= mid) update(p << 1, x, d);
else update(p << 1 | 1, x, d);
tree[p].suf = max(tree[p << 1 | 1].suf, tree[p << 1].suf + tree[p << 1 | 1].sum);
tree[p].sum = tree[p << 1].sum + tree[p << 1 | 1].sum;
tree[p].sum2 = tree[p << 1].sum2 + tree[p << 1 | 1].sum2;
}
tuple2 query(int p, int idx) {
if (tree[p].r <= idx) return tuple2{tree[p].suf, tree[p].sum, tree[p].sum2};
if (idx <= tree[p].l + tree[p].r >> 1) return query(p << 1, idx);
tuple2 ret1 = tuple2{tree[p << 1].suf, tree[p << 1].sum, tree[p << 1].sum2};
tuple2 ret2 = query(p << 1 | 1, idx);
ret1.x += ret2.y, ret1.y = ret2.y = ret1.y + ret2.y, ret1.z = ret2.z = ret1.z + ret2.z;
return ret2.x >= ret1.x ? ret2 : ret1;
}
int query2(int p, long long val) {
if (tree[p].l == tree[p].r) return tree[p].l;
return tree[p << 1].sum2 >= val ? query2(p << 1, val) : query2(p << 1 | 1, val - tree[p << 1].sum2);
}
int main() {
int n = read(), m = read(), q = read();
for (int i = 1; i <= q; ++ i) {
int opt = read();
if (opt == 1) {
int l = read(), r = read();
c[i] = read(), k[i] = read();
vec1[l].push_back(i), vec2[r + 1].push_back(i);
} else if (opt == 2) {
int l = read(), r = read();
k[i] = -read();
vec1[l].push_back(i), vec2[r + 1].push_back(i);
} else {
int a = read();
long long b = readll();
vec3[a].push_back(tuple{b, ++ qstot, i});
}
}
build(1, 1, q);
for (int i = 1; i <= n; ++ i) {
for (int j : vec1[i]) update(1, j, k[j]);
for (int j : vec2[i]) update(1, j, -k[j]);
for (tuple j : vec3[i]) {
tuple2 tmp = query(1, j.z);
if (tmp.x < j.x) continue;
ans[j.y] = c[query2(1, j.x - tmp.x + tmp.z)];
}
}
for (int i = 1; i <= qstot; ++ i) printf("%d\n", ans[i]);
return 0;
}
「JOISC 2019 Day4」蛋糕拼接 3
怎么全是JOI.jpg
容易想到最优情况下就是把 \(C_i\) 从小到大排序围成一圈，美观程度为 \(\sum w-2(\max\{C\}-\min\{C\})\)。
按照 \(C_i\) 排序后，相当于找 \(2(C_i-C_j)+cost(j,i)\) 的最大值，其中 \(cost(l,r)\) 表示 \([l,r]\) 内前 \(m\) 大数之和。
对于每个 \(i\)，\(2(C_i-C_j)+cost(j,i)\) 的最优决策点 \(j\) 具有单调性，于是可以整体二分。
处理那个 \(cost(l,r)\) 的方法有一万种，有莫队套 set，莫队套可删堆，主席树，莫队套树状数组……
莫队套树状数组跑得最快，在经过一些卡常之后竟然以 38ms 的真·优势超过 nodgd 抢到 loj 榜一，泪目
#include <cstdio>
#include <queue>
#include <algorithm>
#define gc (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 65536, stdin), p1 == p2) ? EOF : *p1 ++)
char buf[65536], *p1, *p2;
inline int read() {
char ch;
int x = 0;
while ((ch = gc) < 48);
do x = x * 10 + ch - 48; while ((ch = gc) >= 48);
return x;
}
struct node {
int w, h, rnk;
inline bool operator < (const node a) const {return h < a.h;}
} a[200005];
long long ans = -1e18, c1[200005], sum;
int w[200005], c2[200005], n, m, L = 1, R = 0, N, lim;
inline void insert(int x, int d) {
for (register int i = x; i <= N; i += (i & ~i + 1)) c1[i] += d, ++ c2[i];
}
inline void remove(int x, int d) {
for (register int i = x; i <= N; i += (i & ~i + 1)) c1[i] -= d, -- c2[i];
}
inline long long query() {
register int idx = 0, sum2 = 0;
register long long sum1 = 0;
for (register int i = lim, nxt; i >= 0; -- i) {
nxt = idx | 1 << i;
if (nxt <= N && sum2 + c2[nxt] <= m) sum1 += c1[idx = nxt], sum2 += c2[nxt];
}
return sum1 + (m - sum2) * w[idx + 1];
}
inline long long cost(int l, int r) {
while (L > l) -- L, insert(a[L].rnk, a[L].w);
while (R < r) ++ R, insert(a[R].rnk, a[R].w);
while (L < l) remove(a[L].rnk, a[L].w), ++ L;
while (R > r) remove(a[R].rnk, a[R].w), -- R;
return query();
}
void solve(int l, int r, int x, int y) {
int mid = l + r >> 1, k = 0;
long long now = -1e18;
for (int i = std::min(mid - m + 1, y); i >= x; -- i) {
register long long tmp = 1ll * a[i].h - a[mid].h + cost(i, mid);
if (!k || tmp > now) now = tmp, k = i;
}
ans = std::max(ans, now);
if (l < mid) solve(l, mid - 1, x, k ? k : y);
if (mid < r) solve(mid + 1, r, k ? k : x, y);
}
int main() {
n = read(), m = read();
for (int i = 1; i <= n; ++ i) a[i].w = w[i] = read(), a[i].h = read() << 1;
std::sort(a + 1, a + n + 1);
std::sort(w + 1, w + n + 1);
N = std::unique(w + 1, w + n + 1) - w - 1;
for (int i = 1; i <= n; ++ i) a[i].rnk = N - (std::lower_bound(w + 1, w + N + 1, a[i].w) - w) + 1;
std::reverse(w + 1, w + N + 1);
while (1 << lim + 1 <= N) ++ lim;
solve(m, n, 1, n - m + 1);
printf("%lld", ans);
return 0;
}
