【题目链接】
http://www.lydsy.com/JudgeOnline/problem.php?id=1023
【题解】
仙人掌入门题,拿圆方树练练手。
圆方树就是把一个环建一个新方点,然后向每个在环上的点连边,
接下来就很方便了,用dp的方式求出直径(以这个点为子树的最大直径),在遇到方点时,在环上转两圈。
复杂度 O(m+nm+n)
tips:vector用起来很爽
/* --------------
user Vanisher
problem bzoj-1023
----------------*/
# include <bits/stdc++.h>
# define N 500010
using namespace std;
struct node{
int data,vote;
};
vector <node> e[N];
int use[N],low[N],dfn[N],ti,st[N],top,num,f[N],ans,h[N],po[N],n,m,su[N],sv[N],len,tag[N],place,fx;
int read(){
int tmp=0, fh=1; char ch=getchar();
while (ch<'0'||ch>'9'){if (ch=='-') fh=-1; ch=getchar();}
while (ch>='0'&&ch<='9'){tmp=tmp*10+ch-'0'; ch=getchar(); }
return tmp*fh;
}
void build(int u, int v, int w){
e[u].push_back((node){v,w});
e[v].push_back((node){u,w});
}
void tarjan(int x, int fa){
use[x]=true; low[x]=dfn[x]=++ti;
for (int ed=1; ed<e[x].size(); ed++){
if (dfn[x]<dfn[e[x][ed].data]||e[x][ed].data==fa) continue;
st[++top]=x;
if (use[e[x][ed].data]==false) {
tarjan(e[x][ed].data,x);
if (st[top]==x){
su[++place]=x, sv[place]=e[x][ed].data;
top--;
}
}
else {
int to=e[x][ed].data;
e[++num].push_back((node){0,0});
int s=top;
while (st[top]!=to) top--;
for (int i=top; i<=s; i++)
e[num].push_back((node){st[i],min(s-i+1,i-top)});
tag[num]=s-top+1; top--;
}
}
}
void dp(int x, int fa){
if (tag[x]==0){
int mx1=0, mx2=0;
for (int ed=1; ed<e[x].size(); ed++)
if (e[x][ed].data!=fa){
dp(e[x][ed].data,x);
int now=f[e[x][ed].data]+e[x][ed].vote;
if (now>mx1) mx2=mx1, mx1=now;
else if (now>mx2) mx2=now;
}
ans=max(ans,mx1+mx2);
f[x]=mx1;
}
else {
for (int ed=1; ed<e[x].size(); ed++)
if (e[x][ed].data!=fa){
int t=e[x][ed].data;
dp(e[x][ed].data,x);
f[x]=max(f[e[x][ed].data]+e[x][ed].vote,f[x]);
}
for (int ed=1; ed<e[x].size(); ed++)
if (e[x][ed].data!=fa) h[ed]=f[e[x][ed].data];
else h[ed]=0;
int pl=1, pr=0;
for (int i=1; i<e[x].size(); i++){
while (pl<=pr&&i-po[pl]>tag[x]/2) pl++;
if (pl<=pr) ans=max(ans,h[i]+st[pl]+i-po[pl]);
while (pl<=pr&&h[i]-i>=st[pr]-po[pr]) pr--;
st[++pr]=h[i], po[pr]=i;
}
for (int i=1; i<e[x].size(); i++){
while (pl<=pr&&i+tag[x]-po[pl]>tag[x]/2) pl++;
if (pl<=pr) ans=max(ans,h[i]+st[pl]+i+tag[x]-po[pl]);
while (pl<=pr&&h[i]-i-tag[x]>=st[pr]-po[pr]) pr--;
st[++pr]=h[i], po[pr]=i+tag[x];
}
}
}
int main(){
n=read(), m=read(); num=n; int sum=0;
for (int i=1; i<=n; i++) e[i].push_back((node){0,0});
for (int i=1; i<=m; i++){
int l=read(),la=read();
for (int j=2; j<=l; j++){
int now=read();
build(la,now,1);
la=now;
}
}
tarjan(1,0);
for (int i=1; i<=n; i++) e[i].clear(), e[i].push_back((node){0,0});
for (int i=n+1; i<=num; i++)
for (int j=1; j<e[i].size(); j++)
e[e[i][j].data].push_back((node){i,e[i][j].vote});
for (int i=1; i<=place; i++)
build(su[i],sv[i],1);
dp(1,0);
printf("%d\n",ans);
return 0;
}