洛谷P2921 [USACO08DEC]在农场万圣节Trick or Treat on the Farm

明显一颗内向树。

然后显然内向树里面强连通分量只有可能是环。

于是可以很愉快的写tarjan了2333

最后记忆化统计一下就ok。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <stack>
using namespace std;
const int MAXN = 1e5 + 20;

int N;
int to[MAXN];

namespace SCC
{
    stack<int> sta; bool ins[MAXN];
    int col[MAXN], dfn[MAXN], low[MAXN];
    int cntscc;
    int sum[MAXN];

    void tarjan(int u)
    {
        static int dfsclock = 0;
        dfn[u] = low[u] = ++dfsclock;
        sta.push(u); ins[u] = true;
        if(!dfn[to[u]]){
            tarjan(to[u]);
            low[u] = min(low[u], low[to[u]]);
        }
        else if(ins[to[u]]) low[u] = min(low[u], dfn[to[u]]);
        
        if(dfn[u] == low[u]){
            int tmp; ++cntscc;
            do{
                tmp = sta.top(); sta.pop(); ins[tmp] = false;
                col[tmp] = cntscc;
                ++sum[cntscc];
            }while(tmp != u);
        }
    }
}

int g[MAXN];

int f[MAXN];
int dfs(int u)
{
    if(!g[u]) return SCC::sum[u];
    if(f[u]) return f[u];
    return f[u] = SCC::sum[u] + dfs(g[u]);

}

int main()
{
    using namespace SCC;
    cin>>N;
    for(int i = 1; i <= N; i++) scanf("%d", &to[i]);
    for(int i = 1; i <= N; i++) if(!dfn[i])
        tarjan(i);
    for(int i = 1; i <= N; i++)
        if(col[to[i]] != col[i])
            g[col[i]] = col[to[i]];

    for(int i = 1; i <= N; i++)
        printf("%d\n", dfs(col[i]));
    return 0;
}