C - ThREE
先考虑如果是一条链怎么做，直接模三分类填就行
现在转换到树上，那不妨先对整棵树红蓝交替染色，假设最终红蓝点个数分别为 \(Rnum\) 与 \(Bnum\) ， \(1 \sim n\) 模 3 余 1、2、0 的个数分别为 \(n_1\) 、 \(n_2\) 、 \(n_3\) 。
因此构造染色方法：
- \(Rnum < n_3\) ：红点全放 \(mod \space 3 \equiv 0\) 的点，剩下的数全塞蓝点
- \(Rnum > n_1 + n_3\) ：蓝点全放 \(mod \space 3 \equiv 0\) 的点，剩下的数全塞红点
- \(others\) ：红点放 \(mod \space 3 \equiv 1\) 的点，蓝点放 \(mod \space 3 \equiv 2\) 的点，最后还空着的点就放 \(mod \space 3 \equiv 0\) 的数
#include <bits/stdc++.h>
#define IOS \
std::ios::sync_with_stdio(false); \
std::cin.tie(0); \
std::cout.tie(0);
int main()
{
IOS;
int n;
std::cin >> n;
std::vector<int> p(n + 1);
std::queue<int> q[2]; // 0 红 1 蓝
std::vector<std::vector<int>> g(n + 1);
for (int i = 1, u, v; i < n; ++i)
{
std::cin >> u >> v;
g[u].push_back(v);
g[v].push_back(u);
}
std::function<void(int, int, int)> dfs = [&](int u, int fa, int col)
{
q[col].push(u);
for (auto v : g[u])
if (v != fa)
dfs(v, u, col ^ 1);
};
dfs(1, 0, 0);
int Rnum = q[0].size(), Bnum = q[1].size();
int n1 = (n - 1) / 3 + 1, n2 = (n - 2) / 3 + 1, n3 = n / 3;
if (Rnum < n3)
{
int x = 1;
while (q[0].size() && 3 * x <= n)
{
p[q[0].front()] = 3 * x;
q[0].pop();
++x;
}
while (q[1].size() && 3 * x <= n)
{
p[q[1].front()] = 3 * x;
q[1].pop();
++x;
}
x = 0;
while (q[1].size() && 3 * x + 1 <= n)
{
p[q[1].front()] = 3 * x + 1;
q[1].pop();
++x;
}
x = 0;
while (q[1].size() && 3 * x + 2 <= n)
{
p[q[1].front()] = 3 * x + 2;
q[1].pop();
++x;
}
}
else if (Rnum > n1 + n3)
{
int x = 1;
while (q[1].size() && 3 * x <= n)
{
p[q[1].front()] = 3 * x;
q[1].pop();
++x;
}
while (q[0].size() && 3 * x <= n)
{
p[q[0].front()] = 3 * x;
q[0].pop();
++x;
}
x = 0;
while (q[0].size() && 3 * x + 1 <= n)
{
p[q[0].front()] = 3 * x + 1;
q[0].pop();
++x;
}
x = 0;
while (q[0].size() && 3 * x + 2 <= n)
{
p[q[0].front()] = 3 * x + 2;
q[0].pop();
++x;
}
}
else
{
int x = 0;
while (q[0].size() && 3 * x + 1 <= n)
{
p[q[0].front()] = 3 * x + 1;
q[0].pop();
++x;
}
x = 0;
while (q[1].size() && 3 * x + 2 <= n)
{
p[q[1].front()] = 3 * x + 2;
q[1].pop();
++x;
}
x = 1;
while (q[0].size() && 3 * x <= n)
{
p[q[0].front()] = 3 * x;
q[0].pop();
++x;
}
while (q[1].size() && 3 * x <= n)
{
p[q[1].front()] = 3 * x;
q[1].pop();
++x;
}
}
for (int i = 1; i <= n; ++i)
std::cout << p[i] << " ";
return 0;
}
D. Flowers
设 \(f[i]\) 表示 \(i\) 朵花有几种吃法，因为每次吃要么吃 1 朵红花要么吃 \(k\) 朵白花，因此有转移方程：
\(f[i] = f[i-1] + f[i-k]\)
最后因为问的是区间，所以作个前缀和就好。
#include <bits/stdc++.h>
#define IOS \
std::ios::sync_with_stdio(false); \
std::cin.tie(0); \
std::cout.tie(0);
using ll = long long;
const int N = 1e5;
const int P = 1e9 + 7;
int main()
{
IOS;
int t, k;
std::cin >> t >> k;
std::vector<ll> f(N + 1), s(N + 1);
f[0] = 1;
for (int i = 1; i <= N; ++i)
{
f[i] = f[i - 1] % P;
if (i >= k)
f[i] = (f[i] + f[i - k]) % P;
s[i] = (s[i] + f[i]) % P;
}
for (int i = 1; i <= N; ++i)
s[i] = (s[i - 1] + s[i]) % P;
for (int i = 1, a, b; i <= t; ++i)
{
std::cin >> a >> b;
std::cout << (s[b] - s[a - 1] + P) % P << std::endl;
}
return 0;
}
E. Pillars
和昨天 \(div2\) 的 \(D\) 有点像，都是用数据结构优化 dp 。
设 \(f[i]\) 表示以 \(i\) 为结尾的子序列的最大长度，有转移方程：
\(f[i] = \underset{1 \leq k \leq i-1}{max}f[k] + 1\)
复杂度为 \(O(n^2)\) ，考虑优化
方法和昨天那题差不多，就是先把 \(a\) 离散化后建一棵线段树维护区间最大值，唯一不同的是由于题目给的条件是绝对值，所以要找左右两个区间，大概就是下面这个样子：
时间复杂度 \(O(nlogn)\)
#include <bits/stdc++.h>
#define IOS \
std::ios::sync_with_stdio(false); \
std::cin.tie(0); \
std::cout.tie(0);
#define PLL std::pair<ll, ll>
using ll = long long;
#define lson o << 1
#define rson o << 1 | 1
#define val first
#define id second
struct SegTree
{
const int n;
std::vector<PLL> max;
SegTree(int n) : n(n), max(n << 2 | 1){};
PLL merge(PLL a, PLL b) { return a.val > b.val ? a : b; }
void update(int o, int l, int r, int x, PLL y)
{
if (l == r)
{
max[o] = merge(max[o], y);
return;
}
int mid = (l + r) >> 1;
if (x <= mid)
update(lson, l, mid, x, y);
else
update(rson, mid + 1, r, x, y);
max[o] = merge(max[lson], max[rson]);
}
PLL query(int o, int l, int r, int x, int y)
{
if (x <= l && r <= y)
return max[o];
PLL res = {0, 0};
int mid = (l + r) >> 1;
if (x <= mid)
res = merge(res, query(lson, l, mid, x, y));
if (mid < y)
res = merge(res, query(rson, mid + 1, r, x, y));
return res;
}
};
int main()
{
IOS;
int n, d;
std::cin >> n >> d;
std::vector<ll> a(n + 1), b(n + 1), s(n + 1);
for (int i = 1; i <= n; b[i] = a[i], ++i)
std::cin >> a[i];
std::sort(b.begin() + 1, b.begin() + n + 1);
int nn = std::unique(b.begin() + 1, b.begin() + n + 1) - b.begin() - 1;
// for (int i = 1; i <= n; ++i)
// a[i] = std::lower_bound(b.begin() + 1, b.begin() + nn + 1, a[i]) - b.begin() + 1;
SegTree t(nn);
std::vector<ll> f(n + 1);
for (int i = 1, l, r, pos; i <= n; ++i)
{
l = std::upper_bound(b.begin() + 1, b.begin() + nn + 1, a[i] - d) - b.begin() + 1 - 1;
r = std::lower_bound(b.begin() + 1, b.begin() + nn + 1, a[i] + d) - b.begin() + 1;
PLL res1 = t.query(1, 1, nn, 1, l), res2 = t.query(1, 1, nn, r, nn);
PLL res = res1.val > res2.val ? res1 : res2;
f[i] = res.val + 1, s[i] = res.id;
pos = std::lower_bound(b.begin() + 1, b.begin() + nn + 1, a[i]) - b.begin() + 1;
t.update(1, 1, nn, pos, {f[i], i});
}
int ans = 0, pos = 0;
for (int i = 1; i <= n; ++i)
if (f[i] > ans)
ans = f[i], pos = i;
std::cout << ans << std::endl;
std::vector<int> path;
path.push_back(pos);
while (s[pos])
{
path.push_back(s[pos]);
pos = s[pos];
}
std::reverse(path.begin(), path.end());
for (auto x : path)
std::cout << x << " ";
return 0;
}