BZOJ 1023: [SHOI2008]cactus仙人掌图

1023: [SHOI2008]cactus仙人掌图
思路：先求出非环上的最长链，然后环上用单调队列优化dp。
代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<double, double>
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(x) cerr << #x << " = " << x << "
";
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head

const int N = 5e4 + 5;
vector<int> g[N];
int fa[N], deep[N], dp[N], a[N*2], n, m, t, ans;
int dfn[N], low[N], cnt;
inline void DP(int u, int v) {
    int t = 0;
    for (int i = v; i != u; i = fa[i]) a[++t] = i;
    a[++t] = u;
    for (int i = 1; i <= t; ++i) a[i+t] = a[i];
    deque<int> q;
    for (int i = 1; i <= 2*t; ++i) {
        while(!q.empty() && i-q.front() > t/2) q.pop_front();
        if(!q.empty()) ans = max(ans, dp[a[i]]+dp[a[q.front()]]+i-q.front());
        while(!q.empty() && dp[a[q.back()]]-q.back() <= dp[a[i]]-i) q.pop_back();
        q.push_back(i);
    }
    for (int i = v; i != u; i = fa[i]) {
        dp[u] = max(dp[u], dp[i]+min(deep[i]-deep[u], deep[v]-deep[i]+1));
    }
}
inline void tarjan(int u, int o) {
    fa[u] = o;
    deep[u] = deep[o]+1;
    dfn[u] = low[u] = ++cnt;
    for (int i = 0; i < g[u].size(); ++i) {
        int v = g[u][i];
        if(v == o) continue;
        if(!dfn[v]) {
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
        }
        else low[u] = min(low[u], dfn[v]);
        if(low[v] > dfn[u]) {
            ans = max(ans, dp[u]+dp[v]+1);
            dp[u] = max(dp[u], dp[v]+1);
        }
    }
    for (int i = 0; i < g[u].size(); ++i) {
        int v = g[u][i];
        if(fa[v] != u && dfn[v] > dfn[u]) {
            DP(u, v);
        }
    }
}
int main() {
    while(~scanf("%d %d", &n, &m)) {
        cnt = ans = 0;
        for (int i = 1; i <= m; ++i) {
            scanf("%d", &t);
            for (int i = 1; i <= t; ++i) scanf("%d", &a[i]);
            for (int i = 2; i <= t; ++i) g[a[i-1]].pb(a[i]), g[a[i]].pb(a[i-1]);
        }
        tarjan(1, 0);
        printf("%d
", ans);
        for (int i = 1; i <= n; ++i) g[i].clear(), low[i] = dfn[i] = dp[i] = 0;
    }
    return 0;
}