UVA1146 Now or later(2-sat,4级)_as you must have experienced, instead of landing i-CSDN博客

本文链接：https://blog.csdn.net/nealgavin/article/details/9257475

Description

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.

$\epsfbox{p3211.eps}$

Jim Tarjan, an air-traffic controller, has asked his brother Robert to help him to improve the behavior of the airport.

Input

The input file, that contains all the relevant data, contains several test cases

Each test case is described in the following way. The first line contains the number n of aircraft ( 2 $\le$ n $\le$ 2000). This line is followed by nlines. Each of these lines contains two integers, which represent the early landing time and the late landing time of an aircraft. Note that all times t are such that 0 $\le$ t $\le$ 10⁷.

Output

For each input case, your program has to write a line that conttains the maximal security gap between consecutive landings.

Sample Input

Sample Output

Note: The input file corresponds to Table 1.

Robert's Hints

Optimization vs. Decision

Robert advises you to work on the decision variant of the problem. It can then be stated as follows: Given an integer p, and an instance of the optimization problem, the question is to decide if there is a solution with security gap p or not. Note that, if you know how to solve the decision variant of an optimization problem, you can build a binary search algorithm to find the optimal solution.

On decision

Robert believes that the decision variant of the problem can be modeled as a very particular boolean satisfaction problem. Robert suggests to associate a boolean variable per aircraft stating whether the aircraft is early (variable takes value ``true'') or late (value ``false''). It should then be easy to see that for some aircraft to land at some time has consequences for the landing times of other aircraft. For instance in Table 1 and with a delay of 10, if aircraft A₁ lands early, then aircraft A₃ has to land late. And of course, if aircraft A₃ lands early, then aircraft A₁ has to land late. That is, aircraft A₁ and A₃ cannot both land early and formula (A₁ $\Rightarrow$ ¬A₃) $\wedge$ (A₃ $\Rightarrow$ ¬A₁) must hold.

And now comes Robert's big insight: our problem has a solution, if and only if we have no contradiction. A contradiction being something like A_i $\Leftrightarrow$ ¬A_i.

思路：二分时间间隔，选早点，还是晚点，典型的TWO-SAT问题。

判断有互斥的，就加有向边就行了。

用强连通，判断下就好了。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
const int mm=4000+9;
const int oo=1e7;
const int mn=mm*mm;
int head[mm];
class node
{
  public:int v,next;
}e[mn];
int h[mm][2];
int n;
int dfs_clock,bcc_no,dfn[mm],e_to[mm],stak[mm],top,edge;
int tarjan(int u)
{
  int lowu,lowv,v;
  lowu=dfn[u]=++dfs_clock;
  stak[++top]=u;
  for(int i=head[u];~i;i=e[i].next)
  {
    v=e[i].v;
    if(!dfn[v])
    {
      lowv=tarjan(v);
      lowu=min(lowu,lowv);
    }
    else if(!e_to[v])
      lowu=min(dfn[v],lowu);
  }
  if(lowu==dfn[u])
  { ++bcc_no;
    while(1)
    {v=stak[top--];
     e_to[v]=bcc_no;
     if(v==u)break;
    }
  }
  return lowu;
}
void bcc_find()
{
   memset(dfn,0,sizeof(dfn));
   memset(e_to,0,sizeof(e_to));
   top=dfs_clock=bcc_no=0;
   int z=n+n;
   for(int i=0;i<z;++i)
    if(!dfn[i])tarjan(i);

}
void data()
{
  memset(head,-1,sizeof(head));edge=0;
}
void add(int u,int v)
{
  e[edge].v=v;e[edge].next=head[u];head[u]=edge++;
}
inline int fabs(int x)
{
  if(x<0)return -x;return x;
}
bool jud(int u,int uu,int v,int vv,int x)
{
  if(fabs(h[u][uu]-h[v][vv])<x)
  return 1;
    return 0;
}
bool judge(int mid)
{ data();
  for(int i=0;i<n;++i)
    for(int ii=0;ii<2;++ii)
    for(int j=i+1;j<n;++j)
    for(int jj=0;jj<2;++jj)
    if(jud(i,ii,j,jj,mid))
    add(i+i+ii,j+j+jj^1),add(j+j+jj,i+i+ii^1);
  bcc_find();/**
  puts("eeeeee");
  for(int i=0;i<n+n;++i)
    printf("%d %d\n",i,e_to[i]);
    puts("eeeeee--");*/
  for(int i=0;i<n;++i)
    if(e_to[i+i]==e_to[i+i+1])return 0;
  return 1;
}
int main()
{ int a,b;
  while(~scanf("%d",&n))
  {   int l=0,r=0;
      int ans=0,mid;
     for(int i=0;i<n;++i)
      {scanf("%d%d",&h[i][0],&h[i][1]);
       r=max(r,max(h[i][0],h[i][1]));
      }
      while(l<=r)///二分间隔
      {
         mid=(l+r)/2;
         if(judge(mid)){l=mid+1;ans=max(ans,mid);}
         else r=mid-1;
      }
      printf("%d\n",ans);
  }
  return 0;
}