Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2754 Accepted Submission(s): 918
Robbie has cracked the game, and he has known all the candidate places of each round before the game starts. Now he wants to know the maximum score he can get with the optimal strategy.
给出一些点对,你可以在每对中任意选一个,只能选一个,放置一个炸弹,每个炸弹爆炸时都有一个效果范围,会波及到其放置点为圆心,半径为 r 的圆的范围,问如果要让任意两个圆都不相交(可以相切)的话,半径的最大值是多少。
很裸的 2-SAT 模型,每组点分为 A 和 A‘ ,然后2分枚举半径的值,如果两点间距离小于半径的二倍,那么这两点不能同时放置炸弹,也就是说他们矛盾,根据这个关系建边,判断是否存在解,如果存在说明半径还可以继续增大,否则,半径要减小。
PS:这个题的精度问题,输出要求的是10的负二次方,但是精度只开到 1e-3 会wa
#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int VM=210; const int EM=40010; const double eps=1e-8; struct Edge{ int to,nxt; }edge[EM<<1]; int n,m,cnt,dep,top,atype,head[VM]; int dfn[VM],low[VM],vis[VM],belong[VM]; int stack[VM]; double point[VM][2]; void Init(){ cnt=0, atype=0, dep=0, top=0; memset(head,-1,sizeof(head)); memset(vis,0,sizeof(vis)); memset(low,0,sizeof(low)); memset(dfn,0,sizeof(dfn)); memset(belong,0,sizeof(belong)); } void addedge(int cu,int cv){ edge[cnt].to=cv; edge[cnt].nxt=head[cu]; head[cu]=cnt++; } void Tarjan(int u){ dfn[u]=low[u]=++dep; stack[top++]=u; vis[u]=1; for(int i=head[u];i!=-1;i=edge[i].nxt){ int v=edge[i].to; if(!dfn[v]){ Tarjan(v); low[u]=min(low[u],low[v]); }else if(vis[v]) low[u]=min(low[u],dfn[v]); } int j; if(dfn[u]==low[u]){ atype++; do{ j=stack[--top]; belong[j]=atype; vis[j]=0; }while(u!=j); } } bool CalDis(int i,int j,double mid){ return (point[i][0]-point[j][0])*(point[i][0]-point[j][0])+(point[i][1]-point[j][1])*(point[i][1]-point[j][1])-4*mid*mid<eps; } int solve(double mid){ for(int i=1;i<=n;i++) for(int j=i+1;j<=n;j++){ if(CalDis(i,j,mid)){ addedge(i,j+n); addedge(j,i+n); } if(CalDis(i,j+n,mid)){ addedge(i,j); addedge(j+n,i+n); } if(CalDis(i+n,j,mid)){ addedge(i+n,j+n); addedge(j,i); } if(CalDis(i+n,j+n,mid)){ addedge(i+n,j); addedge(j+n,i); } } for(int i=1;i<=2*n;i++) if(!dfn[i]) Tarjan(i); for(int i=1;i<=n;i++) if(belong[i]==belong[i+n]) return 0; return 1; } int main(){ //freopen("input.txt","r",stdin); while(~scanf("%d",&n)){ for(int i=1;i<=n;i++) scanf("%lf%lf%lf%lf",&point[i][0],&point[i][1],&point[i+n][0],&point[i+n][1]); double l=0,r=100000,mid,ans=0; while(l+eps<=r){ Init(); //因为二分每次都得重新建边,所以初始化在这里 mid=(l+r)/2; if(solve(mid)){ //return true,说明边太少了,应该增大mid,所以l = mid l=mid; ans=mid; //ans=max(ans,mid); //printf("ans=%.2lf\n",ans); }else //return false,说明边太多了,应该减小mid,所以r = mid r=mid; } printf("%.2lf\n",ans); } return 0; }