BZOJ4379: [POI2015]Modernizacja autostrady

画个图意会一下，对于每条边，砍掉它后再把这两棵树接起来，新树的最小直径是两棵树最大直径/2（上取整）的和+1，最大值是直径的和
所以对于每条边都求出它分开的两棵树的直径，找到最小最大值后随便搜一下就行了
emmmmmmmmmmmmmmm
dp的过程细节挺爆炸的

code：

#include<set>#include<map>#include<deque>#include<queue>#include<stack>#include<cmath>#include<ctime>#include<bitset>#include<string>#include<vector>#include<cstdio>#include<cstdlib>#include<cstring>#include<climits>#include<complex>#include<iostream>#include<algorithm>#define ll long long#define lowbit(x) x&(-x)using namespace std;inline void read(int &x){    char c; while(!((c=getchar())>='0'&&c<='9'));    x=c-'0';    while((c=getchar())>='0'&&c<='9') (x*=10)+=c-'0';}inline void up(int &x,const int &y){if(x<y)x=y;}const int maxn = 1100000;int n;vector<int>V[maxn];inline void ins(const int x,const int y){V[x].push_back(y);}int fa[maxn],f[maxn],g[maxn];void dfs(const int x){    f[x]=g[x]=0;    for(int i=0;i<V[x].size();i++)    {        int y=V[x][i]; if(y==fa[x]) continue;        fa[y]=x;        dfs(y);        up(g[x],max(g[y],f[x]+f[y]+1));        up(f[x],f[y]+1);    }}int u[maxn],ug[maxn];int pre[maxn],preg[maxn],suf[maxn],sufg[maxn];int ans1,X1,Y1,ans2,X2,Y2;void dfs2(const int x){    for(int i=0;i<V[x].size();i++)    {        int y=V[x][i];        pre[i+1]=max(pre[i],y==fa[x]?0:f[y]+1),        preg[i+1]=max(preg[i],y==fa[x]?0:max(pre[i]+f[y]+1,g[y]));    }    suf[V[x].size()]=0; sufg[V[x].size()]=0;    for(int i=V[x].size()-1;i>=0;i--)    {        int y=V[x][i];        suf[i]=max(suf[i+1],y==fa[x]?0:f[y]+1),        sufg[i]=max(sufg[i+1],y==fa[x]?0:max(suf[i+1]+f[y]+1,g[y]));    }    for(int i=0;i<V[x].size();i++) if(V[x][i]!=fa[x])    {        int y=V[x][i];        int now=max(pre[i]+suf[i+1],u[x]+max(pre[i],suf[i+1]));         up(now,ug[x]); up(now,preg[i]); up(now,sufg[i+1]);        up(ug[y],now);        int temp=(now+1)/2+(g[y]+1)/2+1;        up(temp,now); up(temp,g[y]);        if(temp<ans1) ans1=temp,Y1=y;        temp=now+g[y]+1;        if(temp>ans2) ans2=temp,Y2=y;        u[y]=max(pre[i],suf[i+1])+1; up(u[y],u[x]+1);    }    for(int i=0;i<V[x].size();i++) if(V[x][i]!=fa[x]) dfs2(V[x][i]);}void trans(const int x){    f[x]=g[x]=0;    for(int i=0;i<V[x].size();i++) if(V[x][i]!=fa[x])    {        int y=V[x][i]; fa[y]=x; trans(y);         up(g[x],g[y]); up(g[x],f[x]+f[y]+1);         up(f[x],f[y]+1);    }}int root,standard;void search(const int x){    bool flag=true;    if(u[x]>standard) flag=false;    for(int i=0;i<V[x].size()&&flag;i++) if(V[x][i]!=fa[x])        if(f[V[x][i]]+1>standard) flag=false;    if(flag) { root=x; return; }    for(int i=0;i<V[x].size();i++)        pre[i+1]=max(pre[i],V[x][i]==fa[x]?0:f[V[x][i]]+1);    suf[V[x].size()]=0;    for(int i=V[x].size()-1;i>=0;i--)        suf[i]=max(suf[i+1],V[x][i]==fa[x]?0:f[V[x][i]]+1);    for(int i=0;i<V[x].size();i++) if(V[x][i]!=fa[x])    {        int y=V[x][i];        u[y]=max(pre[i],suf[i+1])+1; up(u[y],u[x]+1);    }    for(int i=0;i<V[x].size()&&root==-1;i++) if(V[x][i]!=fa[x]) search(V[x][i]);}void search2(const int x){    if(u[x]==standard) { root=x; return; }    for(int i=0;i<V[x].size();i++)        pre[i+1]=max(pre[i],V[x][i]==fa[x]?0:f[V[x][i]]+1);    suf[V[x].size()]=0;    for(int i=V[x].size()-1;i>=0;i--)        suf[i]=max(suf[i+1],V[x][i]==fa[x]?0:f[V[x][i]]+1);    for(int i=0;i<V[x].size();i++) if(V[x][i]!=fa[x])    {        int y=V[x][i];        u[y]=max(pre[i],suf[i+1])+1; up(u[y],u[x]+1);    }    for(int i=0;i<V[x].size()&&root==-1;i++) if(V[x][i]!=fa[x]) search2(V[x][i]);}void solve1(){    printf("%d %d %d ",ans1,X1,Y1);    fa[X1]=Y1; fa[Y1]=X1;    trans(X1); trans(Y1);    root=-1; u[X1]=0; standard=(g[X1]+1)/2; search(X1); printf("%d ",root);    root=-1; u[Y1]=0; standard=(g[Y1]+1)/2; search(Y1); printf("%d\n",root);}void solve2(){    printf("%d %d %d ",ans2,X2,Y2);    fa[X2]=Y2; fa[Y2]=X2;    trans(X2); trans(Y2);    root=-1; u[X2]=0; standard=g[X2]; search2(X2); printf("%d ",root);    root=-1; u[Y2]=0; standard=g[Y2]; search2(Y2); printf("%d\n",root);}int main(){    read(n);    for(int i=1;i<n;i++)    {        int x,y; read(x); read(y);        ins(x,y); ins(y,x);    }    fa[1]=0; dfs(1);    u[1]=0; ans1=2*n,ans2=-1;    dfs2(1); X1=fa[Y1]; X2=fa[Y2];    solve1(); solve2();    return 0;}