链接：
https://vjudge.net/problem/UVALive-3211
题意：
As you must have experienced, instead of landing immediately, an aircraft sometimes waits in a holding
loop close to the runway. This holding mechanism is required by air traffic controllers to space apart
aircraft as much as possible on the runway (while keeping delays low). It is formally defined as a
“holding pattern” and is a predetermined maneuver designed to keep an aircraft within a specified
airspace (see Figure 1 for an example).
Figure 1: A simple Holding Pattern as described in a pilot text book.
Jim Tarjan, an air-traffic controller, has asked his brother Robert to help him to improve the
behavior of the airport.
The TRACON area
The Terminal Radar Approach CONtrol (TRACON) controls aircraft approaching and departing
when they are between 5 and 50 miles of the airport. In this final scheduling process, air traffic
controllers make some aircraft wait before landing. Unfortunately this “waiting” process is complex
as aircraft follow predetermined routes and their speed cannot be changed. To reach some degree of
flexibility in the process, the basic delaying procedure is to make aircraft follow a holding pattern that
has been designed for the TRACON area. Such patterns generate a constant prescribed delay for an
aircraft (see Figure 1 for an example). Several holding patterns may exist in the same TRACON.
In the following, we assume that there is a single runway and that when an aircraft enters the
TRACON area, it is assigned an early landing time, a late landing time and a possible holding pattern.
The early landing time corresponds to the situation where the aircraft does not wait and lands as
soon as possible. The late landing time corresponds to the situation where the aircraft waits in the
prescribed holding pattern and then lands at that time. We assume that an aircraft enters at most
one holding pattern. Hence, the early and late landing times are the only two possible times for the
landing.
The security gap is the minimal elapsed time between consecutive landings. The objective is to
maximize the security gap. Robert believes that you can help.
Example
Assume there are 10 aircraft in the TRACON area. Table 1 provides the corresponding early and
late landing times (columns “Early” and “Late”).
Aircraft Early Late Solution
A1 44 156 Early
A2 153 182 Early
A3 48 109 Late
A4 160 201 Late
A5 55 186 Late
A6 54 207 Early
A7 55 165 Late
A8 17 58 Early
A9 132 160 Early
A10 87 197 Early
Table 1: A 10 aircraft instance of the problem.
The maximal security gap is 10 and the corresponding solution is reported in Table 1 (column
“Solution”). In this solution, the aircraft land in the following order: A8, A1, A6, A10, A3, A9, A2, A7,
A5, A4. The security gap is realized by aircraft A1 and A6.
思路：
先考虑二分，每次重新建图。训练指南上的建图是传入两个或的关系，要满足这个或为真即可满足2-SAT。
此题考虑两个飞机的降落时间不满足条件，加入图中。
代码：
//#include<bits/stdc++.h>
#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<string.h>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<math.h>
#include<stdio.h>
#include<map>
#include<stack>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int MOD = 20071027;
const int MAXN = 2e3+10;
int Next[4][2] = {-1, 0, 0, 1, 1, 0, 0, -1};
vector<int> G[MAXN*2];
stack<int> St;
int dfn[MAXN*2], low[MAXN*2], sccnum[MAXN*2];
int a[MAXN][2];
int dfn_clock, scc_cnt;
int n, m;
void tarjan(int u)
{
dfn[u] = low[u] = ++dfn_clock;
St.push(u);
for (int i = 0;i < (int)G[u].size();i++)
{
int v = G[u][i];
if (!dfn[v])
{
tarjan(v);
low[u] = min(low[v], low[u]);
}
else if (!sccnum[v])
low[u] = min(dfn[v], low[u]);
}
if (low[u] == dfn[u])
{
++scc_cnt;
while(true)
{
int x = St.top();
St.pop();
sccnum[x] = scc_cnt;
if (x == u)
break;
}
}
}
bool solve()
{
memset(dfn, 0, sizeof(dfn));
memset(low, 0, sizeof(low));
memset(sccnum, 0, sizeof(sccnum));
dfn_clock = scc_cnt = 0;
for (int i = 0;i < 2*n;i++)
if (!dfn[i]) tarjan(i);
for (int i = 0;i < 2*n;i+=2)
if (sccnum[i] == sccnum[i+1]) return false;
return true;
}
void add_edge(int x, int xval, int y, int yval)
{
x = x*2+xval;
y = y*2+yval;
G[x^1].push_back(y);
G[y^1].push_back(x);
}
bool check(int mid)
{
for (int i = 0;i < 2*n;i++)
G[i].clear();
for (int i = 0;i < n;i++) for (int p1 = 0;p1 < 2;p1++)
{
for (int j = i+1;j < n;j++) for (int p2 = 0;p2 < 2;p2++)
{
if (abs(a[i][p1]-a[j][p2]) < mid)
add_edge(i, p1^1, j, p2^1);
}
}
return solve();
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
while(cin >> n)
{
int l = 0, r = 0;
for (int i = 0;i < n;i++)
{
cin >> a[i][0] >> a[i][1];
r = max(r, max(a[i][0], a[i][1]));
}
int ans = 0;
while(l <= r)
{
int mid = (l+r)/2;
if (check(mid))
{
ans = max(ans, mid);
l = mid+1;
}
else
r = mid-1;
}
cout << ans << endl;
}
return 0;
}