顯然刪掉的邊肯定是直徑上的邊。考慮枚舉刪哪一條。然后考慮怎么連。顯然新邊應該滿足其兩端點在各自樹中作為根能使樹深度最小。只要線性求出這個東西就可以了，這與求樹的重心的過程類似。

#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long #define N 5010 char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() {int x=0,f=1;char c=getchar();while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();return x*f; } int n,p[N],deep[N],fa[N],f[N],len[N],t,root,ans=N*N; bool flag[N]; struct data{int to,nxt,len; }edge[N<<1]; void addedge(int x,int y,int z){t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].len=z,p[x]=t;} void dfs(int k) {for (int i=p[k];i;i=edge[i].nxt)if (edge[i].to!=fa[k]){deep[edge[i].to]=deep[k]+edge[i].len;fa[edge[i].to]=k;len[edge[i].to]=edge[i].len;dfs(edge[i].to);} } int dp(int k,int ban) {flag[k]=1;int mx=0,mx2=0,ans=0;for (int i=p[k];i;i=edge[i].nxt)if (!flag[edge[i].to]&&edge[i].to!=ban){ans=max(ans,dp(edge[i].to,ban));int x=f[edge[i].to]+edge[i].len;if (x>mx) mx2=mx,mx=x;else if (x>mx2) mx2=x;}f[k]=mx;return max(ans,mx+mx2); } int findroot(int k,int ban,int last) {int mx=0,mx2=0,l=0,len=0;for (int i=p[k];i;i=edge[i].nxt)if (edge[i].to!=fa[k]&&edge[i].to!=ban){int x=f[edge[i].to]+edge[i].len;if (x>mx) mx2=mx,mx=x,l=edge[i].to,len=edge[i].len;else if (x>mx2) mx2=x;}if (max(last,mx2)+len<mx) return findroot(l,ban,max(last,mx2)+len);else return k; } int main() { #ifndef ONLINE_JUDGEfreopen("bzoj4890.in","r",stdin);freopen("bzoj4890.out","w",stdout);const char LL[]="%I64d\n"; #elseconst char LL[]="%lld\n"; #endifn=read();for (int i=1;i<n;i++){int x=read(),y=read(),z=read();addedge(x,y,z),addedge(y,x,z);}dfs(1);int root=1;for (int i=2;i<=n;i++) if (deep[i]>deep[root]) root=i;fa[root]=deep[root]=0;dfs(root);int x=1;for (int i=2;i<=n;i++) if (deep[i]>deep[x]) x=i;while (x!=root){memset(f,0,sizeof(f));memset(flag,0,sizeof(flag));int t=max(dp(root,x),dp(x,fa[x]));int u=findroot(root,x,0),v=findroot(x,fa[x],0);memset(f,0,sizeof(f));memset(flag,0,sizeof(flag));dp(u,x),dp(v,fa[x]);ans=min(ans,max(f[u]+f[v]+len[x],t));x=fa[x];}cout<<ans;return 0; }

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轉載于:https://www.cnblogs.com/Gloid/p/10031368.html

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