luogu P4410 [HNOI2009] no return Island

Face questions

https://www.luogu.org/problemnew/show/P4410

Meaning of the questions: ask the cactus maximum independent set.

1. $ Orzyyb $ $ tarjan $ $ dp $ in the way of thinking is a really wonderful.
2. At the same time deal with two arrays, reducing the complexity of thinking (I wrote the code I jo there is always an aesthetic violence).
3. $ G [x] [0/1] $ represent subtrees $ x $ and $ x $, if x is selected, the largest independent set is.
4. $ F [x] [0/1] [0/1] $ meaningful if and only if $ $ X on the ring, and not for the first ring. $ X $ is provided as a ring head, $ T $ tail ring, $ f [y] [0/1] [0/1] $ $ Y represents a selected or not selected $, $ T $ not selected or selected, $ y $ and $ y $ sub-tree of the largest independent set is.
5. Then transferred on it, pay attention to discounting the effect of previously transferred $ t $ in $ y $ in.

// luogu-judger-enable-o2
#include<cstdio>
#include<cstring>
#include<iostream>
#include<stack>
#include<vector>
#define INF 1000000007
#define ri register int
#define N 100500
using namespace std;

int n,tn,m;
int cir[N];
int a[N];
int ans;

inline int read() {
  int ret=0,f=0; char ch=getchar();
  while (ch<'0' || ch>'9') f|=(ch=='-'),ch=getchar();
  while (ch>='0'&&ch<='9') ret*=10,ret+=ch-'0',ch=getchar();
  return f?-ret:ret;
}

struct graph {
  vector<int> to[N];
  int dfn[N],low[N],cnt;
  stack<int> s;
  int f[N][2][2],g[N][2];
    // state:t  state:x
  void add_edge(int a,int b) {
    to[a].push_back(b);
    to[b].push_back(a);
  }
  void tarjan(int x) {
    dfn[x]=low[x]=++cnt; s.push(x);
    g[x][1]=a[x];
    g[x][0]=0;
    for (ri i=0;i<to[x].size();i++) {
      int y=to[x][i];
      if (dfn[y]) low[x]=min(low[x],dfn[y]);
      else {
        tarjan(y);
        low[x]=min(low[x],low[y]);
        if (low[y]>=dfn[x]) {
          int t=s.top();
                    
          if (t==y) {
             g[x][1]+=max(g[y][0],0);
             g[x][0]+=max(max(g[y][1],g[y][0]),0);
             continue;
          }
                    
          int pret=t;
          s.pop();
                    
          f[t][1][1]=g[t][1];
          f[t][0][0]=g[t][0];
          f[t][1][0]=f[t][0][1]=-INF;
                    
          do {
            t=s.top(); s.pop();
            f[t][1][0]=g[t][1]-max(g[pret][0],0)+max(f[pret][0][0],0);
            f[t][0][0]=g[t][0]-max(max(g[pret][0],g[pret][1]),0)+max(max(f[pret][0][0],f[pret][1][0]),0);
            f[t][1][1]=g[t][1]-max(g[pret][0],0)+max(f[pret][0][1],0);
            f[t][0][1]=g[t][0]-max(max(g[pret][0],g[pret][1]),0)+max(max(f[pret][0][1],f[pret][1][1]),0);
            pret=t;
          }
          while (t!=y);
          g[x][1]+=max(f[y][0][0],0);
          g[x][0]+=max(max(max(f[y][0][0],f[y][0][1]),max(f[y][1][0],f[y][1][1])),0);
        }
        else {
          g[x][1]+=max(g[y][0],0);
          g[x][0]+=max(max(g[y][1],g[y][0]),0);
        }
      }
    }
  }
} G;

int main() {
  tn=n=read(); m=read();
  for (ri i=1;i<=m;i++) {
    int u=read(),v=read();
    G.add_edge(u,v);
  }
  for (ri i=1;i<=n;i++) a[i]=read();
  G.cnt=0;
  G.tarjan(1);
  int ans=0;
  for (ri i=1;i<=n;i++) ans=max(ans,max(G.g[i][0],G.g[i][1]));
  printf("%d\n",ans);
}