HDU-3622 Bomb Game 2sat

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3622

　　题意：一个平面上有很多的炸弹，每个炸弹的爆炸范围是一样的，求最大的爆炸范围使得炸弹之间不相互影响。

　　二分爆炸范围，然后建立2sat模型，看是否存在解。

  1 //STATUS:C++_AC_171MS_972KB

  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=210;

 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 double d[N][N];

 58 int nod[N][2];

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

 60 int n,mt,cnt;

 61 

 62 struct Edge{

 63     int u,v;

 64 }e[N*N*2];

 65 

 66 double dist(int i,int j){

 67     return sqrt((double)((nod[i][0]-nod[j][0])*(nod[i][0]-nod[j][0])+

 68                 (nod[i][1]-nod[j][1])*(nod[i][1]-nod[j][1])));

 69 }

 70 

 71 void adde(int a,int b)

 72 {

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

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

 75 }

 76 

 77 int dfs(int u)

 78 {

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

 80     if(vis[u])return 1;

 81     int i;

 82     vis[u]=1;

 83     S[cnt++]=u;

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

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

 86     }

 87     return 1;

 88 }

 89 

 90 int Twosat()

 91 {

 92     int i,j;

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

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

 95         cnt=0;

 96         if(!dfs(i)){

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

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

 99         }

100     }

101     return 1;

102 }

103 

104 void init(double limt)

105 {

106     int i,j;

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

108     mem(first,-1);

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

110         for(j=i+2;j<n;j++)if(d[i][j]<limt)adde(i,j^1),adde(j,i^1);

111         i++;

112         for(j=i+1;j<n;j++)if(d[i][j]<limt)adde(i,j^1),adde(j,i^1);

113     }

114 }

115 

116 double binary(double l,double r)

117 {

118     double mid;

119     while(fabs(l-r)>EPS){

120         mid=(l+r)/2;

121      //   printf("%.2lf %.2lf %.2lf\n",l,r,mid);

122         init(mid);

123         if(Twosat())l=mid;

124         else r=mid;

125     }

126     return mid;

127 }

128 

129 int main()

130 {

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

132     int i,j;

133     double hig;

134     while(~scanf("%d",&n))

135     {

136         n<<=1;

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

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

139         }

140         hig=0;

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

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

143                 d[i][j]=d[j][i]=dist(i,j);

144                 hig=Max(hig,d[i][j]);

145             }

146         }

147 

148         printf("%.2lf\n",binary(0,hig)/2);

149     }

150     return 0;

151 }