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 }
