Gym 101201F Illumination (Two-Sat)

题意：一个n*n的房子，有很多灯，每个格子只能被上下方向照一次、左右方向照一次，每个灯可以选择上下或是左右照，照明长度以自身位置为中心，占用2*r+1个格子。问能否安排一种方案，使所有格子满足条件。

析：典型的Two-Sat，对于行来说，如果两个能够交叉，那么他们不能都是左右，对于列也是一样。

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(i,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 double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e4 + 50;
const LL mod = 1e9 + 7;
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;
}

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

  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 < n*2; i += 2){
      if(!mark[i] && !mark[i+1]){
        c = 0;
        if(!dfs(i)){
          while(c > 0)  mark[S[--c]] = false;
          if(!dfs(i+1))  return false;
        }
      }
    }
    return true;
  }
};
map<P, int> mp;
vector<int> row[1005], col[1005];
TwoSat twosat;

int main(){
  int r, l;
  scanf("%d %d %d", &n, &r, &l);
  int cnt = 0;
  for(int i = 0; i < l; ++i){
    int x, y;
    scanf("%d %d", &x, &y);
    --x;  --y;
    mp[P(x, y)] = cnt++;
    row[x].pb(y);
    col[y].pb(x);
  }
  for(int i = 0; i < n; ++i){
    sort(row[i].begin(), row[i].end());
    sort(col[i].begin(), col[i].end());
  }
  twosat.n = cnt;
  for(int i = 0; i < n; ++i){
    for(int j = 0; j < row[i].sz; ++j){
      int k = j + 1;
      while(k < row[i].sz && row[i][j] + r >= row[i][k] - r){
        twosat.add_clause(mp[P(i, row[i][j])], 0, mp[P(i, row[i][k])], 0);
        ++k;
      }
    }
    for(int j = 0; j < col[i].sz; ++j){
      int k = j + 1;
      while(k < col[i].sz && col[i][j] + r >= col[i][k] - r){
        twosat.add_clause(mp[P(col[i][j], i)], 1, mp[P(col[i][k], i)], 1);
        ++k;
      }
    }
  }

  puts(twosat.solve() ? "YES" : "NO");
  return 0;
}