bzoj1023 [SHOI2008]cactus仙人掌图

题目描述：

bz

luogu

题解：

仙人掌$tarjan$。

维护这个点向下走的最长链长度。

搞到环的话拆开环用单调栈扫。

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 50050;
const int M = 100050;
template<typename T>
inline void read(T&x)
{
    T f = 1,c = 0;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){c=c*10+ch-'0';ch=getchar();}
    x = f*c;
}
int n,m,hed[N],cnt=1;
struct EG
{
    int to,nxt;
}e[2*M];
void ae(int f,int t)
{
    e[++cnt].to = t;
    e[cnt].nxt = hed[f];
    hed[f] = cnt;
}
void build()
{
    int a,b=-1,c;
    read(a);
    while(a--)
    {
        read(c);
        if(~b)ae(b,c),ae(c,b);
        b=c;
    }
}
int dp[N],dep[N],low[N],tim;
int sta[N],tl,ans;
int r[2*N],tr,s[N],h,l;
void sol()
{
    h = 1,l = 0;
    int lim = tr/2;
    for(int i=1;i<=lim;i++)
    {
        while(h<=l&&dp[r[s[l]]]+s[l]<=dp[r[i]]+i)l--;
        s[++l] = i;r[tr+i]=i;
    }
    for(int i=1;i<=tr;i++)
    {
        while(h<=l&&s[h]<=i)h++;
        while(h<=l&&dp[r[s[l]]]+s[l]<=dp[r[i+lim]]+i+lim)l--;
        s[++l] = i+lim;
        ans = max(ans,dp[r[i]]-i+dp[r[s[h]]]+s[h]);
    }
    for(int i=2;i<=tr;i++)
        dp[r[1]]=max(dp[r[1]],min(i-1,tr+1-i)+dp[r[i]]);
}
void tarjan(int u,int f)
{
    dp[u] = 0;
    dep[u] = low[u] = ++tim;
    sta[++tl] = u;
    for(int j=hed[u];j;j=e[j].nxt)
    {
        int to = e[j].to;
        if(to==f)continue;
        if(!dep[to])
        {
            tarjan(to,u);
            low[u] = min(low[u],low[to]);
            if(low[to]>dep[u])
            {
                ans = max(ans,dp[u]+dp[to]+1);
                dp[u] = max(dp[u],dp[to]+1);
                tl--;
            }else if(low[to]==dep[u])
            {
                r[tr=1]=u;
                while(sta[tl+1]!=to)r[++tr]=sta[tl--];
                sol();
            }
        }else    low[u] = min(low[u],dep[to]);
    }
}
int main()
{
    read(n),read(m);
    while(m--)build();
    tarjan(1,-1);
    printf("%d\n",ans);
    return 0;
}