UVa 3211 Now or later (二分+2-Sat)

时间：2023-03-09 20:15:13
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;

}

　　

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