题解|创世纪题解

这道题的边连起来的话是个基环树森林的样子(i往a[i]连边)，先考虑如果是个树怎么做，设f[i][0/1]表示节点i不选和选可投放的最大节点数量，则（因为一定要有至少一个子节点不选以限制当前节点）。可以先连的有向边，拓扑排序找出每个基环树里面的环，然后断掉环上的一条边来实现这个过程（在这之前重新建一个的有向边的图以方便dp转移）。然后假设断掉的边是，就没有用到p可以限制a[p]的这个条件，强制不选p以限制a[p]，再进行一次树形dp，那么，并用来更新答案即可。总的答案即为各个基环树的答案之和

#include <bits/stdc++.h>
using namespace std;

namespace io {
char buf[1<<21], *p1 = buf, *p2 = buf;
inline char gc() {
    if(p1 != p2) return *p1++;
    p1 = buf;
    p2 = p1 + fread(buf, 1, 1 << 21, stdin);
    return p1 == p2 ? EOF : *p1++;
}
#define G gc

#ifndef ONLINE_JUDGE
#undef G
#define G getchar
#endif

template<class I>
inline void read(I &x) {
    x = 0; I f = 1; char c = G();
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = G(); }
    while(c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = G(); }
    x *= f;
}
template<class I>
inline void write(I x) {
    if(x == 0) {putchar('0'); return;}
    I tmp = x > 0 ? x : -x;
    if(x < 0) putchar('-');
    int cnt = 0;
    while(tmp > 0) {
        buf[cnt++] = tmp % 10 + '0';
        tmp /= 10;
    }
    while(cnt > 0) putchar(buf[--cnt]);
}

#define in(x) read(x)
#define outn(x) write(x), putchar('\n')
#define out(x) write(x), putchar(' ')

} using namespace io;

#define inf 0x3f3f3f3f
#define ll long long
const int N = 1000100;

int c[N], Ctot;
int n, a[N], f[N][2], in[N], vis[N];
// f[i][0] 不投放i， f[I][1] 投放i 
int cnt, head[N];
struct edge {
    int to, nxt;
} e[N << 1];

void ins(int u, int v) {
    e[++cnt] = (edge) {v, head[u]};
    head[u] = cnt;
}

void col(int u, int bl) {
    c[u] = bl;
    for(int i = head[u]; i; i = e[i].nxt) if(!c[e[i].to]) col(e[i].to, bl);
}

int q[N], p;
void topsort() {
    int l = 1, r = 1;
    for(int i = 1; i <= n; ++i) if(!in[i]) q[r++] = i;
    while(l < r) {
        int u = q[l++];
        for(int i = head[u]; i; i = e[i].nxt) {
            int v = e[i].to; in[v]--;
            if(!in[v]) q[r++] = v;
        }
    }
}

void dfs(int u, int fa) {
    f[u][0] = 0;
    int num = inf;
    for(int i = head[u]; i; i = e[i].nxt) {
        int v = e[i].to;
//        if(v == fa) continue;
        if(v == p && u == a[p]) continue;
//        if(v == a[p] && u == p) continue;
        dfs(v, u);
        f[u][0] += max(f[v][0], f[v][1]);
        num = min(num, max(f[v][0], f[v][1]) - f[v][0]);
    }
    f[u][1] = f[u][0] - num + 1;
}

void dfs1(int u, int fa) {
    f[u][0] = 0;
    int num = inf;
    for(int i = head[u]; i; i = e[i].nxt) {
        int v = e[i].to;
//        if(v == fa) continue;
        if(v == p && u == a[p]) continue;
//        if(v == a[p] && u == p) continue;
        dfs1(v, u);
        f[u][0] += max(f[v][0], f[v][1]);
        num = min(num, max(f[v][0], f[v][1]) - f[v][0]);
    }
    if(u == a[p]) f[u][1] = f[u][0] + 1;
    else f[u][1] = f[u][0] + 1 - num;
//    printf("test %d %d\n", u, a[p]);
}

int main() {
    read(n);
    for(int i = 1; i <= n; ++i) {
        read(a[i]); in[a[i]]++;
        ins(i, a[i]);
    }
    topsort();
    memset(head, 0, sizeof(head));
    cnt = 0;
    for(int i = 1; i <= n; ++i) ins(a[i], i);
    for(int i = 1; i <= n; ++i) if(!c[i]) col(i, ++Ctot);
    int ans = 0;
    for(int i = 1; i <= n; ++i) {
        if(in[i] && !vis[c[i]]) {
            vis[c[i]] = 1; p = i;
            dfs(p, p);
//            for(int j = 1; j <= n; ++j) printf("%d %d\n", f[j][0], f[j][1]);
//            puts("");
            int sum = max(f[p][0], f[p][1]);
            dfs1(p, p);
//            for(int j = 1; j <= n; ++j) printf("%d %d\n", f[j][0], f[j][1]);
//            puts("");
            sum = max(sum, f[p][0]);
            ans += sum;
        }
    }
    outn(ans);
}