codeforces 698b 图论

题意： 给一串数字a1~an,ai表示i与ai相连，问给出的数组经过最少数量的更改能形成一颗树

题解：分析可知组成的只能由三种非连通图构成，点，不含环链，含环链，含环链必定只可能还有一个环

　　　统计三种的数量，点和不含环链必定会有父节点为自身的点，含环链用强连通跑一跑，最后把这三种图连到一个节点上就行

　　　注意一下只含有含环链的情况

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <map>
#include <vector>
#include <queue>
using namespace std;
#define EPS 1e-10
typedef long long ll;
const int maxn = 2e5 + 10;
const int INF = 1e9 ;
const int N = maxn;
const int M = maxn;

struct Edge {
    int to, next;
} edge[M*2];
int head[N];
int cnt_edge;

void add_edge(int u, int v)
{
    edge[cnt_edge].to = v;
    edge[cnt_edge].next = head[u];
    head[u] = cnt_edge;
    cnt_edge++;
}
vector<int>q;
int dfn[N];
int low[N];
int stk[N];
bool in[N];
int kind[N];

int top;
int idx;
int cnt;

int n, m,num;

void dfs(int u)
{
    dfn[u] = low[u] = ++idx;
    in[u] = true;
    stk[++top] = u;
    for (int i = head[u]; i != -1; i = edge[i].next)
    {
        int v = edge[i].to;
        if (!dfn[v])
        {
            dfs(v);
            low[u] = min(low[v], low[u]);
        }
        else if(in[v])
        {
            low[u] = min(low[u], dfn[v]);
        }
    }

    if (low[u] == dfn[u])
    {
        ++cnt;
        int j;
        num = 0;
        do {
            num++;
            j = stk[top--];
            in[j] = false;
            kind[j] = cnt;
        } while (j != u);
        if(num > 1)
        q.push_back(stk[top+1]);
    }
}

void init()
{
    memset(dfn, 0, sizeof dfn);
    memset(head, -1, sizeof head);
    cnt_edge = 0;
    top = cnt = idx = 0;
}
int main(){
   // freopen("in.txt","r",stdin);
    cin>>n;
    int a[maxn];
    init();
    int cnt = 0;
    for(int i = 1;i <= n;i++){
        scanf("%d",&a[i]);
        if(i != a[i]){
            add_edge(i,a[i]);
        }
        else{ q.push_back(i);
            cnt = 1;
        }
    }

    for(int i = 1;i <= n;i++){
    if(!dfn[i]){
        dfs(i);
    }
    }

    if(num == n){
        cout<<1<<endl;
        a[q[0]] = q[0];
        for(int i = 1;i <= n;i++)
            if(i == n) printf("%d
",a[i]);
            else printf("%d ",a[i]);
    }
    else{
        if(cnt)
            cout<<q.size() - 1<<endl;
        else
            cout<<q.size()<<endl;
        for(int i = 0;i < q.size();i++)
            a[q[i]] = q[0];
        for(int i = 1;i <= n;i++){
            if(i == n) printf("%d
",a[i]);
            else printf("%d ",a[i]);
        }
    }
    return 0;
}