POJ-2296 Map Labeler 2sat

　　题目链接：http://poj.org/problem?id=2296

　　二分+2sat，每个点的上下两个方向为2sat的两个状态。

  1 //STATUS:C++_AC_16MS_536KB

  2 #include <functional>

  3 #include <algorithm>

  4 #include <iostream>

  5 //#include <ext/rope>

  6 #include <fstream>

  7 #include <sstream>

  8 #include <iomanip>

  9 #include <numeric>

 10 #include <cstring>

 11 #include <cassert>

 12 #include <cstdio>

 13 #include <string>

 14 #include <vector>

 15 #include <bitset>

 16 #include <queue>

 17 #include <stack>

 18 #include <cmath>

 19 #include <ctime>

 20 #include <list>

 21 #include <set>

 22 #include <map>

 23 using namespace std;

 24 //using namespace __gnu_cxx;

 25 //define

 26 #define pii pair<int,int>

 27 #define mem(a,b) memset(a,b,sizeof(a))

 28 #define lson l,mid,rt<<1

 29 #define rson mid+1,r,rt<<1|1

 30 #define PI acos(-1.0)

 31 //typedef

 32 typedef long long LL;

 33 typedef unsigned long long ULL;

 34 //const

 35 const int N=110;

 36 const int INF=0x3f3f3f3f;

 37 const int MOD=5000,STA=100010;

 38 const LL LNF=1LL<<60;

 39 const double EPS=1e-8;

 40 const double OO=1e15;

 41 const int dx[4]={-1,0,1,0};

 42 const int dy[4]={0,1,0,-1};

 43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

 44 //Daily Use ...

 45 inline int sign(double x){return (x>EPS)-(x<-EPS);}

 46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

 47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

 48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

 49 template<class T> inline T Min(T a,T b){return a<b?a:b;}

 50 template<class T> inline T Max(T a,T b){return a>b?a:b;}

 51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

 52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

 53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

 54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

 55 //End

 56 

 57 int nod[N][2];

 58 int first[N*2],next[N*N*4],vis[N*2],S[N*2];

 59 int T,n,mt,cnt;

 60 

 61 struct Edge{

 62     int u,v;

 63 }e[N*N*4];

 64 

 65 void adde(int a,int b)

 66 {

 67     e[mt].u=a,e[mt].v=b;

 68     next[mt]=first[a];first[a]=mt++;

 69 }

 70 

 71 int dfs(int u)

 72 {

 73     if(vis[u^1])return 0;

 74     if(vis[u])return 1;

 75     int i;

 76     vis[u]=1;

 77     S[cnt++]=u;

 78     for(i=first[u];i!=-1;i=next[i]){

 79         if(!dfs(e[i].v))return 0;

 80     }

 81     return 1;

 82 }

 83 

 84 int Twosat()

 85 {

 86     int i,j;

 87     for(i=0;i<n;i+=2){

 88         if(vis[i] || vis[i^1])continue;

 89         cnt=0;

 90         if(!dfs(i)){

 91             while(cnt)vis[S[--cnt]]=0;

 92             if(!dfs(i^1))return 0;

 93         }

 94     }

 95     return 1;

 96 }

 97 

 98 int judge(double *a1,double *a2,double *b1,double *b2)

 99 {

100     if(Max(a1[0],a2[0])<=Min(b1[0],b2[0])

101        || Min(a1[0],a2[0])>=Max(b1[0],b2[0])

102        || Max(a1[1],a2[1])<=Min(b1[1],b2[1])

103        || Min(a1[1],a2[1])>=Max(b1[1],b2[1]))return 0;

104     return 1;

105 }

106 

107 void init(double limt)

108 {

109     int i,j,x,y;

110     double a1[2],a2[2],b1[2],b2[2];

111     mt=0;mem(vis,0);

112     mem(first,-1);

113     for(i=0;i<n;i++){

114         for(j=i+1;j<n;j++){

115             x=i<<1;y=j<<1;

116             a1[0]=nod[i][0]-limt/2,a1[1]=nod[i][1];b1[0]=nod[j][0]-limt/2,b1[1]=nod[j][1];

117             a2[0]=nod[i][0]+limt/2,b2[0]=nod[j][0]+limt/2;

118             a2[1]=nod[i][1]-limt;b2[1]=nod[j][1]-limt;

119             if(judge(a1,a2,b1,b2)){

120                 adde(x,y^1);adde(y,x^1);

121             }

122             b2[1]=nod[j][1]+limt;

123             if(judge(a1,a2,b1,b2)){

124                 adde(x,y);adde(y^1,x^1);

125             }

126             a2[1]=nod[i][1]+limt;b2[1]=nod[j][1]-limt;

127             if(judge(a1,a2,b1,b2)){

128                 adde(x^1,y^1);adde(y,x);

129             }

130             b2[1]=nod[j][1]+limt;

131             if(judge(a1,a2,b1,b2)){

132                 adde(x^1,y);adde(y^1,x);

133             }

134         }

135     }

136 }

137 

138 int binary(int l,int r)

139 {

140     int mid;

141     while(l<r){

142         mid=(l+r)>>1;

143      //   printf("%d %d\n",l,r);

144         init(mid);

145         if(Twosat())l=mid+1;

146         else r=mid;

147     }

148     return l;

149 }

150 

151 int main()

152 {

153  //   freopen("in.txt","r",stdin);

154     int i,j;

155     scanf("%d",&T);

156     while(T--)

157     {

158         scanf("%d",&n);

159         for(i=0;i<n;i++){

160             scanf("%d%d",&nod[i][0],&nod[i][1]);

161         }

162 

163         printf("%d\n",binary(0,20001)-1);

164     }

165     return 0;

166 }