CF1268E Happy Cactus

https://www.luogu.com.cn/problem/CF1268E

就这还3400？？？

首先考虑一棵树的情况，设一开始\(f[i]=1\)

把边按边权从大到小插入，假设插入边\((u,v,i)\)

显然\(f[u]=f[v]=f[u]+f[v]\)

考虑仙人掌的情况，先考虑一个环，无非就是链接最小边的时候，最大边两端的会被多算一次，减掉即可

设\(g[i]=f[u]+f[v]\)

那么就是\(f[u]=f[v]=f[u]+f[v]-g[max(C)], g[i]=f[u]+f[v]\)
\(max(C)\)表示当前环上的最大边

code:

	#include<bits/stdc++.h>
#define N 1000050
using namespace std;
struct edge {
    int v, nxt, c;
} e[N << 1];
int p[N], eid;
void init() {
    memset(p, -1, sizeof p);
    eid = 0;
}
void insert(int u, int v, int c) {
    e[eid].v = v;
    e[eid].c = c;
    e[eid].nxt = p[u];
    p[u] = eid ++;
}

int c[N], gs, sz, col[N], mi[N], ma[N], n, m;
int calc() {
    if(gs == 1) return 0;    
    ma[sz] = 0, mi[sz] = m + 1;
    int x = 0, y = 0;
    for(int i = 1; i <= gs; i ++) {
        if(c[i] > ma[sz]) ma[sz] = c[i], x = i;
        if(c[i] < mi[sz]) mi[sz] = c[i], y = i;
    }

    for(int i = 1; i <= gs; i ++) c[i + gs] = c[i];
    int l = x, r = y;
    if(l > r) r += gs;
    for(int i = l; i < r; i ++) if(c[i] < c[i + 1]) return 0;

    l = x, r = y;
    if(l < r) l += gs;
    for(int i = l; i > r; i --) if(c[i] < c[i - 1]) return 0;
    return 1;    
}
int vis[N], fa[N], a[N], ok[N];
void dfs(int u) {
    vis[u] = 1;
    for(int i = p[u]; i + 1; i = e[i].nxt) {
        int v = e[i].v, w = e[i].c;
        if(v == fa[u]) continue;
        if(!vis[v]) {
            fa[v] = u, a[v] = w;
            dfs(v);
        } else if(vis[v] == 2) {
            gs = 0;
            ++ sz;
            int x = v;            
            while(x != u) c[++ gs] = a[x], col[a[x]] = sz, x = fa[x];
            col[w] = sz; c[++ gs] = w;
            ok[sz] = calc();
        }
    }
    vis[u] = 2;
}
int f[N], g[N], hau[N], hav[N];
int main() {
    init();
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; i ++) {
        int u, v;
        scanf("%d%d", &u, &v);
        insert(u, v, i), insert(v, u, i);
        hau[i] = u, hav[i] = v;
    }
    //printf("114514");
    

    dfs(1);
    for(int i = 1; i <= n; i ++) f[i] = 1;
    for(int i = m; i >= 1; i --) {
        int u = hau[i], v = hav[i];
        g[i] = f[u] = f[v] = f[u] + f[v] - (ok[col[i]] && i == mi[col[i]]) * g[ma[col[i]]];
    }
    for(int i = 1; i <= n; i ++) printf("%d ", f[i] - 1);
    return 0;
}