好久沒有2-sat了，此題當復習之用，二分求最大值+2-sat判斷可行，此題主要跪于題意：The results should be rounded to three decimal
places. You should promise that there is still no overlap for any two balloons after rounded.
rounded是四舍五入的意思，按要求，3位之后要全部舍去，知道：printf（）,自動四舍五入。所有-0.0005

之后四舍五入即為要求。

#include<iostream> #include<queue> #include<stack> #include<cstdio> #include<vector> using namespace std; int n; struct points {int x,y,z; }; points point[402]; inline int getdis(points a,points b) //三維 {return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z); } int vis[402];int dfn[402];int low[402];int scc[402];int numscc=0; int times=0; stack<int>sta;int ins[402]; vector<vector<int> >v(402); void clear() {for(int i=0;i<2*n;i++){ins[i]=vis[i]=dfn[i]=low[i]=scc[i]=0;v[i].clear();}numscc=times=0; } void tarjian(int u) {dfn[u]=low[u]=++times;sta.push(u);ins[u]=1;int len=v[u].size();for(int i=0;i<len;i++){int uu=v[u][i];if(!vis[uu]){vis[uu]=1;tarjian(uu);if(low[uu]<low[u])low[u]=low[uu];}else if(ins[uu]&&dfn[uu]<low[u])low[u]=dfn[uu];}if(low[u]==dfn[u]){int cur;numscc++;do{cur=sta.top();sta.pop();ins[cur]=0;scc[cur]=numscc;}while(cur!=u);} } bool check(double r) {clear();for(int i=0;i<2*n-2;i++)for(int j=(i%2==0?i+2:i+1);j<2*n;j++){double dis=0.00+getdis(point[i],point[j]);if(dis<4*r*r){v[i].push_back(j^1);v[j].push_back(i^1);}}for(int i=0;i<2*n;i++){if(!vis[i]){vis[i]=1;tarjian(i);}}for(int i=0;i<2*n;i+=2){if(scc[i]==scc[i+1])return 0;}return 1; } int main() {while(~scanf("%d",&n)){for(int i=0;i<2*n;i++)scanf("%d%d%d",&point[i].x,&point[i].y,&point[i].z);double l=0,r=100000,mid;while(r-l>0.000001) //二分。{mid=(r+l)/2;if(check(mid))l=mid;elser=mid;}double ll=l-0.0005;printf("%.3lf\n",ll);}return 0; }

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