P3119 [USACO15JAN]草鉴定Grass Cownoisseur 分层图或者跑两次最长路

https://www.luogu.org/problemnew/show/P3119

题意

有一个有向图，允许最多走一次逆向的路，问从1再走回1，最多能经过几个点。

思路

（一）
首先先缩点。自己在缩点再建图中犯了错误，少连接了大点到其他点的边。
跑两次最长路，一次以1为起点，一次以1为终点（跑一遍反图）
然后枚举边，判断可否形成一个环。
（二）
分层图的思想
以为只有一次逆向的机会，可以建两层图，第一层向第二层连翻转的情况。

#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>

using namespace std;
#define lson (l, mid, rt << 1)
#define rson (mid + 1, r, rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define boost                    \
    ios::sync_with_stdio(false); \
    cin.tie(0)
#define rep(a, b, c) for (int a = (b); a <= (c); ++a)
#define max3(a, b, c) max(max(a, b), c);
#define min3(a, b, c) min(min(a, b), c);

const ll oo = 1ll << 17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9 + 7;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399; //黄金分割点
const double tPHI = 0.38196601;

template <typename T>
inline T read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x = f ? -x : x;
}

inline void cmax(int &x, int y) {
    if (x < y) x = y;
}
inline void cmax(ll &x, ll y) {
    if (x < y) x = y;
}
inline void cmin(int &x, int y) {
    if (x > y) x = y;
}
inline void cmin(ll &x, ll y) {
    if (x > y) x = y;
}

/*-----------------------showtime----------------------*/
const int maxn = 1e5 + 9;
vector<int> mp1[maxn], mp2[2][maxn];
int dfn[maxn], low[maxn], vis[maxn], gtot, nn, dp[maxn];
int col[maxn], vv[maxn];
stack<int> st;

void tarjan(int u) {
    dfn[u] = low[u] = ++gtot;
    st.push(u);
    vis[u] = 1;
    for (int i = 0; i < mp1[u].size(); i++) {
        int v = mp1[u][i];
        if (dfn[v] == 0) {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        } else if (vis[v]) {
            low[u] = min(low[u], dfn[v]);
        }
    }
    if (low[u] == dfn[u]) {
        int x;
        nn++;
        while (!st.empty()) {
            int x = st.top();
            st.pop();
            col[x] = nn;
            vis[x] = 0;
            dp[nn]++;
            if (x == u) break;
        }
    }
}
int ans = 0;
int dis[maxn][2];
pii edge[maxn];
void dji(int s, int id) {
    priority_queue<pii> que;

    dis[s][id] = dp[s];
    que.push(pii(dis[s][id], s));
    while (!que.empty()) {
        int u = que.top().se;
        que.pop();

        for (int i = 0; i < mp2[id][u].size(); i++) {
            int v = mp2[id][u][i], w = dp[v];

            if (dis[v][id] < dis[u][id] + w) {
                dis[v][id] = dis[u][id] + w;
                que.push(pii(dis[v][id], v));
            }
        }
    }
}
int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    rep(i, 1, m) {
        int x, y;
        scanf("%d%d", &x, &y);
        mp1[x].pb(y);
        edge[i] = pii(x, y);
    }
    for (int i = 1; i <= n; i++)
        if (!dfn[i]) tarjan(i);

    int s = col[1];
    int pp = 0;
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= n; i++) {
        int u = col[i];
        if (u == 0) continue;

        for (int j = 0; j < mp1[i].size(); j++) {
            int v = col[mp1[i][j]];
            if (v == 0 || u == v) continue;

            mp2[0][u].pb(v);
            mp2[1][v].pb(u);
        }
    }
    dji(s, 0);

    dji(s, 1);

    int ans = dp[s];
    for (int i = 1; i <= m; i++) {
        int u = col[edge[i].fi], v = col[edge[i].se];
        if (dis[v][0] && dis[u][1]) ans = max(ans, dis[v][0] + dis[u][1] - dp[s]);
    }

    printf("%d\n", ans);
    return 0;
}