题目链接：https://www.lydsy.com/JudgeOnline/problem.php?id=1040

题目链接：https://www.luogu.com.cn/problem/P2607

题目链接：https://loj.ac/problem/10162

做法

看似有向边，其实可以转化为无向边。

构成一个基环森林

但是需要特判一种情况：当有两个人互相讨厌的时候，形成一棵树，同时有重边（我的找环做法需要sort判重，可能有更好的方法）

然后把森林里的每一棵树或者基环树DP一下（就是没有上司的舞会模型，超水），把答案加起来

注意会爆int

做完了

代码

#include <iostream>
#include <cstdio>
#include <algorithm>
#define Re register
#define LL long long
using namespace std;
inline int read() {
    Re int x = 0;
    char ch = getchar();
    while (!isdigit(ch)) ch = getchar();
    while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
    return x;
}
const int maxN = 1000005;
struct Edge {
    int nxt, to;
} e[maxN << 1];
struct EE {
    int u, v;
} E[maxN];
int cnte = 1, head[maxN];

inline void add_Edge(int i, int j) {
    e[++cnte].nxt = head[i];
    e[cnte].to = j;
    head[i] = cnte;
}
LL f[maxN][2];
int N, val[maxN], huan[maxN], cnt, a, b;
bool vis[maxN], ishuan[maxN], flag;
void dfs(int u, int fa) {
    vis[u] = true;
    for (Re int i = head[u]; i; i = e[i].nxt) {
        Re int v = e[i].to;
        if (v == fa)
            continue;
        if (vis[v]) {
            a = u, b = v;
            continue;
        }
        dfs(v, u);
    }
}
void DP(int u, int fa) {
    f[u][0] = 0, f[u][1] = val[u];
    for (Re int i = head[u]; i; i = e[i].nxt) {
        Re int v = e[i].to;
        if (v == fa || (u == a && v == b) || (v == a && u == b))
            continue;
        DP(v, u);
        f[u][0] += max(f[v][1], f[v][0]);
        f[u][1] += f[v][0];
    }
}
LL Ans = 0;
bool cmp(EE x, EE y) {
    if (x.u != y.u)
        return x.u < y.u;
    return x.v < y.v;
}
int main() {
    N = read();
    for (Re int x, i = 1; i <= N; ++i) {
        val[i] = read(), x = read();
        E[i].u = min(i, x), E[i].v = max(i, x);
    }
    sort(E + 1, E + N + 1, cmp);
    for (int i = 1; i <= N; ++i) {
        if (E[i].u == E[i - 1].u && E[i].v == E[i - 1].v)
            continue;
        add_Edge(E[i].u, E[i].v);
        add_Edge(E[i].v, E[i].u);
    }
    LL res = 0;
    for (int u = 1; u <= N; ++u) {
        if (vis[u])
            continue;
        a = 0, b = 0, dfs(u, 0);
        if (!a || !b) {
            DP(u, 0);
            Ans += max(f[u][1], f[u][0]);
            continue;
        }
        res = 0;
        DP(a, b);
        res = max(res, f[a][0]);
        DP(b, a);
        res = max(res, f[b][0]);
        Ans += res;
    }
    printf("%lld\n", Ans);
    return 0;
}