AtCoder Regular Contest 115 F Migration
这种最大值最小化的题一般可以先考虑二分。设二分了一个 \(mid\)。
记 \(A = (a_1,a_2,...,a_k)\) 为表示每个棋子的位置的状态，如果 \(A,B\) 可以互相到达，就在它们之间连一条无向边。则要判断的是 \(S = (s_1,s_2,...,s_k)\) 和 \(T = (t_1,t_2,...,t_k)\) 是否在同一连通块内。
记 \(f(A)\) 为 \(\sum\limits_{i=1}^k a_{A_i}\)，那我们不妨让 \(S\) 和 \(T\) 都到达一个 \(f(A)\) 最小的状态 \(A\)（如果 \(f(A)\) 相同则比较字典序）。如果两个 \(A\) 都相同，则 \(S,T\) 在同一连通块内，否则不在。
设 \(g(A)\) 为 \(A\) 能到达的最小状态，则问题转化成了求 \(g(A)\)。
考虑一个暴力做法：每次找到能移的棋子中造成的贡献最大的，移动这个棋子（需要预处理点 \(u\) 经过权值不超过 \(D\) 的点能到达的所有点的点权最小值）。因为每个棋子的移动不会重复经过某个点，所以最多移动 \(O(nk)\) 次。时间复杂度是 \(O(nk^2 \log n \log ans)\)，显然会 T。
考虑加点优化。用堆维护下一个贡献最大的操作。每次 \(\sum\limits_{i=1}^k a_i\) 变小时，一些点的移动会变得合法，将它们加入堆中即可。需要精细实现，复杂度 \(O(\text{可过})\)。
code
// Problem: F - Migration
// Contest: AtCoder - AtCoder Regular Contest 115
// URL: https://atcoder.jp/contests/arc115/tasks/arc115_f
// Memory Limit: 1024 MB
// Time Limit: 4000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;
namespace IO {
char ibuf[(1 << 20) + 1], *iS, *iT;
#if ONLINE_JUDGE
#define gh() (iS == iT ? iT = (iS = ibuf) + fread(ibuf, 1, (1 << 20) + 1, stdin), (iS == iT ? EOF : *iS++) : *iS++)
#else
#define gh() getchar()
#endif
inline long long read() {
char ch = gh();
long long x = 0;
bool t = 0;
while (ch < '0' || ch > '9') t |= ch == '-', ch = gh();
while (ch >= '0' && ch <= '9') x = (x << 1) + (x << 3) + (ch ^ 48), ch = gh();
return t ? ~(x - 1) : x;
}
template <typename _Tp>
inline void write(_Tp x) {
static char stk[64], *top = stk;
if (x < 0) {
x = ~(x - 1);
putchar('-');
}
do *top++ = x % 10, x /= 10;
while (x);
while (top != stk) putchar((*--top) | 48);
}
}
using IO::read;
using IO::write;
const int maxn = 2020;
const ll inf = 0x3f3f3f3f3f3f3f3fLL;
ll n, m, a[maxn], head[maxn], len, lsh[maxn], tot;
int c[maxn], d[maxn], p[maxn], e[maxn], to[maxn << 1], nxt[maxn << 1];
pii f[maxn][maxn];
bool vis[maxn];
int stk[maxn];
struct piece {
ll s, t;
} b[maxn];
inline void add_edge(int u, int v) {
to[++len] = v;
nxt[len] = head[u];
head[u] = len;
}
struct node {
int u;
ll val;
int t, i, j;
} h[maxn];
unordered_set<int> st[maxn];
inline bool operator < (const node &a, const node &b) {
if (a.val != b.val) {
return a.val < b.val;
}
bool x = (a.t < a.u), y = (b.t < b.u);
if (x && y) {
return a.i > b.i;
} else if (!x && !y) {
return a.i < b.i;
} else {
return !x;
}
}
struct wwh {
ll x;
int y, z;
} g[maxn * maxn];
inline bool operator < (const wwh &a, const wwh &b) {
return a.x > b.x;
}
namespace SGT {
node tree[maxn << 2];
void build(int rt, int l, int r) {
if (l == r) {
tree[rt] = h[c[l]];
tree[rt].i = l;
return;
}
int mid = (l + r) >> 1;
build(rt << 1, l, mid);
build(rt << 1 | 1, mid + 1, r);
tree[rt] = max(tree[rt << 1], tree[rt << 1 | 1]);
}
void update(int rt, int l, int r, int x, node &y) {
if (l == r) {
tree[rt] = y;
tree[rt].i = l;
return;
}
int mid = (l + r) >> 1;
(x <= mid) ? update(rt << 1, l, mid, x, y) : update(rt << 1 | 1, mid + 1, r, x, y);
tree[rt] = max(tree[rt << 1], tree[rt << 1 | 1]);
}
}
inline void work(ll x) {
ll s = 0;
for (int i = 1; i <= n; ++i) {
st[i].clear();
}
for (int i = 1; i <= m; ++i) {
s += a[c[i]];
st[c[i]].insert(i);
}
int tt = 0;
for (int i = 1; i <= n; ++i) {
node mx = (node){i, -1, 0, i, -1};
for (int j = 1; j <= tot; ++j) {
if (f[i][j].scd == -1) {
continue;
}
node t = (node){i, a[i] - f[i][j].fst, f[i][j].scd, i, j};
if (lsh[j] + s - a[i] <= x) {
p[i] = j;
mx = max(mx, t);
} else if (x + a[i] - lsh[j] > 0 && mx < t) {
g[++tt] = (wwh){x + a[i] - lsh[j], i, j};
}
}
h[i] = mx;
}
sort(g + 1, g + tt + 1);
SGT::build(1, 1, m);
int pos = 1;
while (1) {
node u = SGT::tree[1];
if (u.val < 0 || (u.val == 0 && c[u.i] <= u.t)) {
break;
}
s -= u.val;
st[u.u].erase(u.i);
c[u.i] = u.t;
st[u.t].insert(u.i);
SGT::update(1, 1, m, u.i, h[u.t]);
int top = 0;
while (pos <= tt && s <= g[pos].x) {
int i = g[pos].y, j = g[pos].z;
p[i] = j;
node t = (node){i, a[i] - f[i][j].fst, f[i][j].scd, i, j};
if (h[i] < t) {
h[i] = t;
if (!vis[i]) {
vis[i] = 1;
stk[++top] = i;
}
}
++pos;
}
for (int _ = 1; _ <= top; ++_) {
int i = stk[_];
vis[i] = 0;
for (int x : st[i]) {
SGT::update(1, 1, m, x, h[i]);
}
}
}
}
inline bool check(ll x) {
for (int i = 1; i <= m; ++i) {
c[i] = b[i].s;
}
work(x);
for (int i = 1; i <= m; ++i) {
d[i] = c[i];
c[i] = b[i].t;
}
work(x);
for (int i = 1; i <= m; ++i) {
if (c[i] != d[i]) {
return 0;
}
}
return 1;
}
void dfs(int u, int fa, int mx, int rt) {
f[rt][mx] = min(f[rt][mx], make_pair(a[u], 1LL * u));
for (int i = head[u]; i; i = nxt[i]) {
int v = to[i];
if (v == fa) {
continue;
}
dfs(v, u, max(mx, e[v]), rt);
}
}
void solve() {
n = read();
for (int i = 1; i <= n; ++i) {
a[i] = read();
lsh[++tot] = a[i];
}
sort(lsh + 1, lsh + tot + 1);
tot = unique(lsh + 1, lsh + tot + 1) - lsh - 1;
for (int i = 1; i <= n; ++i) {
e[i] = lower_bound(lsh + 1, lsh + tot + 1, a[i]) - lsh;
}
for (int i = 1, u, v; i < n; ++i) {
u = read();
v = read();
add_edge(u, v);
add_edge(v, u);
}
m = read();
ll s1 = 0, s2 = 0;
for (int i = 1; i <= m; ++i) {
b[i].s = read();
b[i].t = read();
s1 += a[b[i].s];
s2 += a[b[i].t];
}
for (int u = 1; u <= n; ++u) {
for (int i = 0; i <= tot; ++i) {
f[u][i] = make_pair(inf, -1);
}
dfs(u, -1, e[u], u);
for (int i = 1; i <= tot; ++i) {
f[u][i] = min(f[u][i], f[u][i - 1]);
}
}
ll l = max(s1, s2), r = *max_element(a + 1, a + n + 1) * m, ans = -1;
while (l <= r) {
ll mid = (l + r) >> 1;
if (check(mid)) {
ans = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
write(ans);
}
int main() {
solve();
return 0;
}
