Title: https://www.luogu.org/problemnew/show/P1522
Too lazy to carefully divide the situation and write floyd directly like the problem solution and then don't understand the meaning of the last step max...
How to ensure that they are considered separately in one? What if the min of the new connected edge and the max of the original diameter are in three connected blocks?
code show as below:
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> using namespace std; double const inf=0x3f3f3f3f; int n,xx[155],yy[155],col[155],cr; double dis[155][155],ans,mx[155],as,s[155]; double cal(int i,int j) { return sqrt(1.0*(xx[i]-xx[j])*(xx[i]-xx[j])+1.0*(yy[i]-yy[j])*(yy[i]-yy[j])); } //void dfs(int x) //{ // col[x]=cr; // for(int i=1;i<=n;i++) // if(i!=x&&dis[i][x]!=inf&&!col[i])dfs(i); //} int main() { scanf("%d",&n); ans=inf; for(int i=1;i<=n;i++) scanf("%d%d",&xx[i],&yy[i]); int d; for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) { scanf("%1d",&d);// if(d==1)dis[i][j]=cal(i,j); else if(i!=j)dis[i][j]=inf; } // for(int i=1;i<=n;i++) // if(!col[i])cr++,dfs(i); for(int k=1;k<=n;k++) for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]); for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) if(dis[i][j]!=inf) { mx[i]=max(mx[i],dis[i][j]); as=max(as,dis[i][j]); } for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) if(dis[i][j]==inf)ans=min(ans,mx[i]+mx[j]+cal(i,j)); printf("%.6lf",max(ans,as)); return 0; }