1 #include<cstdio> 2 #include<iostream> 3 #define M 400009 4 #define ll long long 5 using namespace std; 6 ll d[M],v[M],ans1; 7 int n,cnt=1,head[M],next[M],u[M],mx,f[M],q[M],ans,mx1,fro[M],from[M],sum[M]; 8 void jia(int a1,int a2,int a3) 9 { 10 cnt++; 11 from[cnt]=a1; 12 next[cnt]=head[a1]; 13 head[a1]=cnt; 14 u[cnt]=a2; 15 v[cnt]=(ll)(a3)*(ll)(n); 16 return; 17 } 18 void bfs(int a1) 19 { 20 d[a1]=0; 21 int h=0,t=1; 22 q[1]=a1; 23 f[a1]=1; 24 fro[a1]=0; 25 sum[a1]=0; 26 for(;h<t;) 27 { 28 int p=q[++h]; 29 for(int i=head[p];i;i=next[i]) 30 if(!f[u[i]]) 31 { 32 f[u[i]]=1; 33 q[++t]=u[i]; 34 d[u[i]]=d[p]+v[i]; 35 fro[u[i]]=i; 36 sum[u[i]]=sum[p]+1; 37 } 38 } 39 return; 40 } 41 int main() 42 { 43 scanf("%d",&n); 44 for(int i=1;i<n;i++) 45 { 46 int a1,a2,a3; 47 scanf("%d%d%d",&a1,&a2,&a3); 48 jia(a1,a2,a3); 49 jia(a2,a1,a3); 50 } 51 bfs(1); 52 ll max1=0; 53 for(int i=1;i<=n;i++) 54 { 55 if(max1<d[i]) 56 { 57 max1=d[i]; 58 mx1=i; 59 } 60 f[i]=0; 61 } 62 bfs(mx1); 63 max1=0; 64 for(int i=1;i<=n;i++) 65 { 66 if(max1<d[i]) 67 { 68 max1=d[i]; 69 ans=i; 70 } 71 f[i]=0; 72 } 73 ans1=d[ans]; 74 printf("%lld\n",d[ans]/n); 75 for(;ans;) 76 { 77 int p=fro[ans]; 78 v[p]--; 79 v[p^1]--; 80 ans=from[p]; 81 } 82 bfs(1); 83 max1=0; 84 for(int i=1;i<=n;i++) 85 { 86 if(max1<d[i]) 87 { 88 max1=d[i]; 89 mx1=i; 90 } 91 f[i]=0; 92 } 93 bfs(mx1); 94 max1=0; 95 for(int i=1;i<=n;i++) 96 if(max1<d[i]) 97 { 98 max1=d[i]; 99 ans=i; 100 } 101 printf("%lld\n",ans1-d[ans]); 102 return 0; 103 }
先找一条直径，把直径上的边的权值减去1，再找一遍直径，差便是答案。
