AT2336 [ARC069D] Flags
AT2336 [ARC069D] Flags
二分答案,2-sat
判定,线段树优化建边。
线段树上父亲向儿子连边。
对于每个点 ,连向与他距离不超过 的点的反点,表示选 就只能选与他距离不超过 的点。
最后跑 tarjan
判定即可。
时间复杂度 。
#include<bits/stdc++.h>
using namespace std;
//#define int long long
typedef long long ll;
#define ha putchar(' ')
#define he putchar('\n')
inline int read() {
int x = 0, f = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (c == '-')
f = -1;
c = getchar();
}
while (c >= '0' && c <= '9')
x = x * 10 + c - '0', c = getchar();
return x * f;
}
inline void write(int x) {
if (x < 0) {
putchar('-');
x = -x;
}
if (x > 9)
write(x / 10);
putchar(x % 10 + '0');
}
const int _ = 8e4 + 10;
int n, cnt, idx, scc, low[_], dfn[_], id[_], Id[_];
stack<int> s;
vector<pair<int, int>> d;
vector<int> e[_];
void build(int o, int l, int r) {
id[o] = ++cnt;
if (o > 1) e[id[o >> 1]].emplace_back(cnt);
if (l == r) {
int v = d[l - 1].second;
e[id[o]].push_back(v <= n ? v + n : v - n);
return;
}
int mid = (l + r) >> 1;
build(o << 1, l, mid), build(o << 1 | 1, mid + 1, r);
}
void upd(int o, int l, int r, int L, int R, int x) {
if(L > R) return;
if (L <= l && r <= R) {
e[x].emplace_back(id[o]);
return;
}
int mid = (l + r) >> 1;
if (L <= mid) upd(o << 1, l, mid, L, R, x);
if (R > mid) upd(o << 1 | 1, mid + 1, r, L, R, x);
}
pair<int, int> get(int i, int m) {
pair<int, int> ret;
int l = 1, r = i, mid;
while (l <= r) {
mid = (l + r) >> 1;
if (d[i - 1].first - d[mid - 1].first >= m) l = mid + 1;
else r = mid - 1;
}
ret.first = r + 1, l = i, r = n << 1;
while (l <= r) {
mid = (l + r) >> 1;
if (d[mid - 1].first - d[i - 1].first < m) l = mid + 1;
else r = mid - 1;
}
ret.second = l - 1;
return ret;
}
void tarjan(int u) {
low[u] = dfn[u] = ++idx;
s.push(u);
for (int v : e[u])
if (!dfn[v]) {
tarjan(v);
low[u] = min(low[u], low[v]);
} else if (!Id[v]) low[u] = min(low[u], dfn[v]);
if (low[u] == dfn[u]) {
++scc;
while (1) {
int nw = s.top();
s.pop();
Id[nw] = scc;
if (nw == u) break;
}
}
}
bool check(int lim) {
scc = idx = 0;
memset(low, 0, sizeof low);
memset(dfn, 0, sizeof dfn);
memset(Id, 0, sizeof Id);
for (int i = 0; i < _; ++i) e[i].clear();
build(1, 1, cnt = n << 1);
for (int i = 1; i <= n << 1; ++i) {
int r = d[i - 1].second;
pair<int, int> p = get(i, lim);
upd(1, 1, n << 1, p.first, i - 1, r);
upd(1, 1, n << 1, i + 1, p.second, r);
}
for (int i = 1; i <= n << 1; ++i)
if (!dfn[i]) tarjan(i);
for (int i = 1; i <= n; ++i)
if (Id[i] == Id[i + n]) return 0;
return 1;
}
signed main() {
n = read();
for (int i = 1, x, y; i <= n; ++i) {
x = read(), y = read();
d.emplace_back(make_pair(x, i));
d.emplace_back(make_pair(y, i + n));
}
sort(d.begin(), d.end());
int l = 0, r = 1e9, mid;
while (l <= r) {
mid = (l + r) >> 1;
if (check(mid)) l = mid + 1;
else r = mid - 1;
}
write(l - 1), he;
return 0;
}
本文来自博客园,作者:蒟蒻orz,转载请注明原文链接:https://www.cnblogs.com/orzz/p/18121999