标签:blog io os ar for sp div on log
算是很裸的吧,二分半径,然后2-sat判定能不能满足条件。
#include <bits/stdc++.h> using namespace std; const int maxn=1222; const double eps=1e-9; struct node{ double x,y; }p[maxn]; struct twoset{ int n; vector<int>g[maxn<<1]; bool mark[maxn<<1]; int s[maxn<<1],c; bool dfs(int x){ if(mark[x^1])return false; if(mark[x])return true; mark[x]=true; s[c++]=x; for(int i=0;i<g[x].size();i++)if(!dfs(g[x][i]))return false; return true; } void init(int n){ this->n=n; for(int i=0;i<n*2;i++)g[i].clear(); memset(mark,0,sizeof mark); } void add_clause(int x,int xval,int y,int yval){ x=x*2+xval; y=y*2+yval; g[x^1].push_back(y); g[y^1].push_back(x); } bool solve(){ for(int i=0;i<n*2;i+=2){ if(!mark[i]&&!mark[i+1]){ c=0; if(!dfs(i)){ while(c>0)mark[s[--c]]=false; if(!dfs(i+1))return false; } } } return true; } }solver; double getdis(node a,node b){ return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); } void run(int n){ for(int i=0;i<n;i++)scanf("%lf%lf",&p[i].x,&p[i].y); double l=0,r=30000,ans=0; while(l+eps<=r){ double mid=(l+r)/2; solver.init(n); for(int i=0;i<n;i++) for(int j=i+1;j<n;j++){ double dis=getdis(p[i],p[j]); if(dis<2*mid){ solver.add_clause(i,0,j,0); solver.add_clause(i,1,j,1); } } if(solver.solve())ans=mid,l=mid+eps; else r=mid-eps; } solver.init(n); for(int i=0;i<n;i++) for(int j=i+1;j<n;j++){ double dis=getdis(p[i],p[j]); if(dis<2*ans){ solver.add_clause(i,0,j,0); solver.add_clause(i,1,j,1); } } solver.solve(); printf("%.8lf\n",ans); for(int i=0;i<n;i++){ if(solver.mark[i<<1])putchar(‘1‘); else putchar(‘2‘); printf("%c",i==n-1?‘\n‘:‘ ‘); } } int main() { // freopen("in","r",stdin); int n; while(scanf("%d",&n)>0)run(n); return 0; }
标签:blog io os ar for sp div on log
原文地址:http://www.cnblogs.com/wshh/p/4029828.html