UVALive - 3211 - Now or later（图论——2-SAT）

Posted 2021-02-17 npugen

Problem   UVALive - 3211 - Now or later

Time Limit: 9000 mSec

Problem Description

 

Input

 

Output

 

Sample Input

10 44 156 153 182 48 109 160 201 55 186 54 207 55 165 17 58 132 160 87 197

Sample Output

10

 

题解：2-SAT问题板子题，这个问题主要是理论难度比较大，有了结论之后代码很容易，有专门的论文阐释算法的正确性，看了几位大佬写的，基本上明白是怎么一回事，理解不深刻，就不在这里胡扯了，直接上代码。

 

  1 #include <bits/stdc++.h>
  2 
  3 using namespace std;
  4 
  5 #define REP(i, n) for (int i = 1; i <= (n); i++)
  6 #define sqr(x) ((x) * (x))
  7 
  8 const int maxn = 2000 + 10;
  9 const int maxm = 30 + 10;
 10 const int maxs = 10000 + 10;
 11 
 12 typedef long long LL;
 13 typedef pair<int, int> pii;
 14 typedef pair<double, double> pdd;
 15 
 16 const LL unit = 1LL;
 17 const int INF = 0x3f3f3f3f;
 18 const LL mod = 1000000007;
 19 const double eps = 1e-14;
 20 const double inf = 1e15;
 21 const double pi = acos(-1.0);
 22 
 23 struct TwoSAT
 24 {
 25     int n;
 26     vector<int> G[maxn * 2];
 27     bool mark[maxn * 2];
 28     int S[maxn * 2], c;
 29 
 30     bool dfs(int x)
 31     {
 32         if (mark[x ^ 1])
 33             return false;
 34         if (mark[x])
 35             return true;
 36         mark[x] = true;
 37         S[c++] = x;
 38         for (auto v : G[x])
 39         {
 40             if (!dfs(v))
 41                 return false;
 42         }
 43         return true;
 44     }
 45 
 46     void init(int n)
 47     {
 48         this->n = n;
 49         for (int i = 0; i < n * 2; i++)
 50         {
 51             G[i].clear();
 52         }
 53         memset(mark, 0, sizeof(mark));
 54     }
 55 
 56     void add_clause(int x, int xval, int y, int yval)
 57     {
 58         x = x * 2 + xval;
 59         y = y * 2 + yval;
 60         G[x ^ 1].push_back(y);
 61         G[y ^ 1].push_back(x);
 62     }
 63 
 64     bool solve()
 65     {
 66         for (int i = 0; i < n * 2; i += 2)
 67         {
 68             if (!mark[i] && !mark[i + 1])
 69             {
 70                 c = 0;
 71                 if (!dfs(i))
 72                 {
 73                     while (c > 0)
 74                     {
 75                         mark[S[--c]] = false;
 76                     }
 77                     if (!dfs(i + 1))
 78                         return false;
 79                 }
 80             }
 81         }
 82         return true;
 83     }
 84 };
 85 
 86 TwoSAT solver;
 87 
 88 int n, T[maxn][2];
 89 
 90 bool Judge(int lim)
 91 {
 92     solver.init(n);
 93     for (int i = 0; i < n; i++)
 94     {
 95         for (int a = 0; a < 2; a++)
 96         {
 97             for (int j = i + 1; j < n; j++)
 98             {
 99                 for (int b = 0; b < 2; b++)
100                 {
101                     if (abs(T[i][a] - T[j][b]) < lim)
102                     {
103                         solver.add_clause(i, a ^ 1, j, b ^ 1);
104                     }
105                 }
106             }
107         }
108     }
109     return solver.solve();
110 }
111 
112 main()
113 {
114     ios::sync_with_stdio(false);
115     cin.tie(0);
116     //freopen("input.txt", "r", stdin);
117     //freopen("output.txt", "w", stdout);
118     while (cin >> n && n)
119     {
120         int le = 0, ri = 0;
121         for (int i = 0; i < n; i++)
122         {
123             for (int j = 0; j < 2; j++)
124             {
125                 cin >> T[i][j];
126                 ri = max(ri, T[i][j]);
127             }
128         }
129 
130         int ans = 0;
131         while (le <= ri)
132         {
133             int mid = (le + ri) >> 1;
134             if (Judge(mid))
135             {
136                 ans = mid;
137                 le = mid + 1;
138             }
139             else
140             {
141                 ri = mid - 1;
142             }
143         }
144         cout << ans << endl;
145     }
146     return 0;
147 }