hdu 3622 bomb game

hdu 3622 bomb game
two_sat
N组，每组2个坐标，选择一个放置炸弹
，范围为半径为ri的圆 要求是任意的
两个圆不能相交，N小于等于100 最后的
得分是N个圆的最小半径 现在求的是最大
得分容易想到二分答案 并进行可行性分析
那么算法就有了。
1：确定上下界、
2：二分
3：判断是否可行
4：在精度范围内则跳出

最终二分的解就是答案

 #include<iostream>#include<cstring>#include<algorithm>#include<cstdio>#include<vector>#include<sstream>#include<string>#include<climits>#include<stack>#include<set>#include<bitset>#include<cmath>#include<deque>#include<map>#include<queue>#define iinf 2000000000#define linf 1000000000000000000LL#define dinf 1e200#define eps 1e-11#define all(v) (v).begin(),(v).end()#define sz(x)  x.size()#define pb push_back#define mp make_pair#define lng long long#define sqr(a) ((a)*(a))#define pii pair<int,int>#define pll pair<lng,lng>#define pss pair<string,string>#define pdd pair<double,double>#define X first#define Y second#define pi 3.14159265359#define ff(i,xi,n) for(int i=xi;i<=(int)(n);++i)#define ffd(i,xi,n) for(int i=xi;i>=(int)(n);--i)#define ffl(i,r) for(int i=head[r];i!=-1;i=edge[i].next)#define cc(i,j) memset(i,j,sizeof(i))#define two(x)((lng)1<<(x))#define N 500#define M 1000000#define lson l , mid , rt << 1#define rson mid + 1 , r , rt << 1 | 1#define Mod  n#define Pmod(x) (x%Mod+Mod)%Modusing namespace std;typedef vector<int>  vi;typedef vector<string>  vs;typedef unsigned int uint;typedef unsigned lng ulng;template<class T> inline void checkmax(T &x,T y){if(x<y) x=y;}template<class T> inline void checkmin(T &x,T y){if(x>y) x=y;}template<class T> inline T Min(T x,T y){return (x>y?y:x);}template<class T> inline T Max(T x,T y){return (x<y?y:x);}template<class T> T gcd(T a,T  b){return (a%b)==0?b:gcd(b,a%b);}template<class T> T lcm(T a,T b){return a*b/gcd(a,b);}template<class T> T Abs(T a){return a>0?a:(-a);}template<class T> inline T lowbit(T n){return (n^(n-1))&n;}template<class T> inline int countbit(T n){return (n==0)?0:(1+countbit(n&(n-1)));}template<class T> inline bool isPrimeNumber(T n){if(n<=1)return false;for (T i=2;i*i<=n;i++) if (n%i==0) return false;return true;}template<class T> inline T Minmin(T a,T b,T c,T d){return Min(Min(a,b),Min(c,d));}struct tarjan_seg{    int head[N],tot,id[N],pre[N],low[N],stack[N],top,n,m,Index,newid,newhead[N],*cur;    struct pp   {        int v,next;    }edge[N*N];    void add(int u,int v)    {        edge[tot].v=v;        edge[tot].next=cur[u];        cur[u]=tot++;    }    void init()    {        cc(head,-1);        tot=0;        newid=0;        cur=head;        cc(pre,-1);        cc(low,-1);        top=0;        cc(id,-1);        Index=0;    }    void tarjan(int r,int p)    {        bool xx=false;        pre[r]=low[r]=++Index;        stack[++top]=r;        ffl(i,r)        {            int v=edge[i].v;            if(v==p&&xx==0)            {                xx=1;                continue;            }            if(pre[v]==-1)            {                tarjan(v,r);                checkmin(low[r],low[v]);            }            else            if(id[v]==-1)            checkmin(low[r],pre[v]);        }        if(pre[r]==low[r])        {            int v;            ++newid;            do            {                v=stack[top--];                id[v]=newid;            }while(v!=r);        }    }    void solve()    {        ff(i,1,n)        if(pre[i]==-1)        tarjan(i,-1);    }    void build_map()    {        int  hash[N]={};        cur=newhead;        ff(i,1,n)        {            ffl(j,i)            {                int v=edge[j].v;                if(id[i]!=id[v]&&hash[id[i]]!=id[j])                add(id[i],id[j]),hash[id[i]]=id[j];            }        }    }    bool two_sat()    {        for(int i=1;i<=n;i+=2)        if(id[i]==id[i+1])        return false;        return true;    }};int nround;int  x[N],y[N];double dis[N][N];double Distance(int i,int j){    return sqrt((double)sqr(x[j]-x[i])+(double)sqr(y[j]-y[i]));} tarjan_seg G;bool check(double r){    G.n=2*nround;    G.init();    ff(i,1,2*nround)    ff(j,1,2*nround)    {        int u=(i+1)/2,v=(j+1)/2;        if(v==u) continue;        if(dis[i][j]<=2*r)        G.add(i,4*v-1-j),G.add(j,4*u-1-i);    }    G.solve();    return G.two_sat();}int main(){    while(scanf("%d",&nround)==1)    {        ff(i,1,2*nround)        {            scanf("%d%d",&x[i],&y[i]);        }        double le=0,ri=-100000,mid;        ff(i,1,2*nround)        ff(j,i+1,2*nround)        {            if((i+1)/2==(j+1)/2) continue;            dis[i][j]=dis[j][i]=Distance(i,j);            checkmax(ri,dis[i][j]);        }        while(ri-le>=eps)        {            mid=(le+ri)/2;            if(check(mid))          le=mid;          else          ri=mid;        }        printf("%.2f\n",Max(le,ri));    }    return 0;}