UVa 3211 Now or later (二分+2-Sat)

题意：有 n 架飞机，每个飞机早着陆，或者晚着陆，让你安排一个方式，让他们着陆的时间间隔尽量大。

析：首先对于时间间隔，可以用二分来解决，然后就成了一个判定性问题，然后怎么判断该时间间隔是不是成立呢，那么用2-Sat能解决，每次对于时间间隔都小于正在判定的，然后给他们连上相应的边，是连两条，然后跑一遍2-sat 就OK了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e15;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2000 + 10;
const int mod = 3;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

// UVaLive 3211

int ta[maxn], tb[maxn];

struct TwoSAT{
  int n;
  vector<int> G[maxn<<1];
  bool mark[maxn<<1];
  int S[maxn<<1], c;

  bool dfs(int x){
    if(mark[x^1])  return false;
    if(mark[x])  return true;
    mark[x] = 1;
    S[c++] = x;
    for(int i = 0; i < G[x].sz; ++i)
      if(!dfs(G[x][i]))  return false;
    return true;
  }

  void init(int n){
    this->n = n;
    for(int i = 0; i < n*2; ++i)  G[i].cl;
    ms(mark, 0);
  }

  void add_clause(int x, int xval, int y, int yval){
    x = x * 2 + xval;
    y = y * 2 + yval;
    G[x^1].pb(y);
    G[y^1].pb(x);
  }

  bool solve(){
    for(int i = 0; i < 2*n; i += 2){
      if(!mark[i] && !mark[i+1]){
        c = 0;
        if(!dfs(i)){
          while(c > 0)  mark[S[--c]] = 0;
          if(!dfs(i+1))  return false;
        }
      }
    }
    return true;
  }
};

TwoSAT twosat;

bool judge(int m){
  twosat.init(n);
  for(int i = 0; i < n; ++i)
    for(int j = i+1; j < n; ++j){
      if(abs(ta[i] - ta[j]) <= m)  twosat.add_clause(i, 0, j, 0);
      if(abs(ta[i] - tb[j]) <= m)  twosat.add_clause(i, 0, j, 1);
      if(abs(tb[i] - ta[j]) <= m)  twosat.add_clause(i, 1, j, 0);
      if(abs(tb[i] - tb[j]) <= m)  twosat.add_clause(i, 1, j, 1);
    }
  return twosat.solve();
}

int main(){
  while(scanf("%d", &n) == 1){
    int l = 0, r = 0;
    for(int i = 0; i < n; ++i){
      scanf("%d %d", ta+i, tb+i);
      r = max(r, tb[i]);
    }
    while(l <= r){
      int m = l + r >> 1;
      if(judge(m))  l = m + 1;
      else r = m - 1;
    }
    printf("%d\n", l);
  }
  return 0;
}