[仙人掌直径 单调队列 DP] BZOJ 1023 [SHOI2008]cactus仙人掌图

题解：http://z55250825.blog.163.com/blog/static/150230809201412793151890/

http://hzwer.com/4645.html

#include
#include
#include
using namespace std;
typedef long long ll;

inline char nc()
{
	static char buf[100000],*p1=buf,*p2=buf;
	if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
	return *p1++;
}

inline void read(int &x)
{
	char c=nc(),b=1;
	for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
	for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

const int N=50005;
const int M=200005;

struct edge{
	int u,v,next;
};

edge G[M];
int head[N],inum=1;

inline void add(int u,int v,int p){
	G[p].u=u; G[p].v=v; G[p].next=head[u]; head[u]=p;
}

int n,m,ans;

int clk,pre[N],low[N];
int fat[N],depth[N];
int f[N];

int a[N<<1],Q[N<<1],l,r;

void dp(int rt,int x)
{
	int pnt=depth[x]-depth[rt]+1;
	for(int i=x;i!=rt;i=fat[i]) a[pnt--]=f[i]; a[pnt]=f[rt];
	pnt=depth[x]-depth[rt]+1;
	for(int i=1;i<=pnt;i++) a[i+pnt]=a[i];
	l=1; r=0;
	for(int i=1;i<=pnt+pnt;i++)
	{
		while(l<=r && i-Q[l]>pnt/2) l++;
		if (l<=r) ans=max(ans,a[i]+i+a[Q[l]]-Q[l]);
		while(l<=r && a[Q[r]]-Q[r]<=a[i]-i) r--;
		Q[++r]=i;
	}
	for(int i=2;i<=pnt;i++)
		f[rt]=max(f[rt],a[i]+min(i-1,pnt-i+1));
}

#define V G[p].v

void dfs(int u,int fa)
{
	pre[u]=low[u]=++clk; depth[u]=depth[fa]+1; fat[u]=fa;
	for (int p=head[u];p;p=G[p].next)
		if (V!=fa)
		{
			if (!pre[V]){
				dfs(V,u);
				low[u]=min(low[u],low[V]);
			}
			else
				low[u]=min(low[u],pre[V]);
			if (low[V]>pre[u]){
				ans=max(ans,f[u]+f[V]+1);
				f[u]=max(f[u],f[V]+1);
			}
		}
	for (int p=head[u];p;p=G[p].next)
		if (u!=fat[V] && pre[u]