2015 NOIP day2 t2 信息传递 tarjan

信息传递

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://www.luogu.org/problem/show?pid=2661

Description

有n个同学（编号为1到n）正在玩一个信息传递的游戏。在游戏里每人都有一个固定的信息传递对象，其中，编号为i的同学的信息传递对象是编号为Ti同学。
　　游戏开始时，每人都只知道自己的生日。之后每一轮中，所有人会同时将自己当前所知的生日信息告诉各自的信息传递对象（注意：可能有人可以从若干人那里
获取信息，但是每人只会把信息告诉一个人，即自己的信息传递对象）。当有人从别人口中得知自己的生日时，游戏结束。请问该游戏一共可以进行几轮？

Under two situations the player could score one point.

⋅1. If you touch a buoy before your opponent, you will get one point. For example if your opponent touch the buoy #2 before you after start, he will score one point. So when you touch the buoy #2, you won't get any point. Meanwhile, you cannot touch buoy #3 or any other buoys before touching the buoy #2.

⋅2. Ignoring the buoys and relying on dogfighting to get point.
If you and your opponent meet in the same position, you can try to
fight with your opponent to score one point. For the proposal of game
balance, two players are not allowed to fight before buoy #2 is touched by anybody.

There are three types of players.

Speeder:
As a player specializing in high speed movement, he/she tries to avoid
dogfighting while attempting to gain points by touching buoys.
Fighter:
As a player specializing in dogfighting, he/she always tries to fight
with the opponent to score points. Since a fighter is slower than a
speeder, it's difficult for him/her to score points by touching buoys
when the opponent is a speeder.
All-Rounder: A balanced player between Fighter and Speeder.

There will be a training match between Asuka (All-Rounder) and Shion (Speeder).
Since the match is only a training match, the rules are simplified: the game will end after the buoy #1 is touched by anybody. Shion is a speed lover, and his strategy is very simple: touch buoy #2,#3,#4,#1 along the shortest path.

Asuka is good at dogfighting, so she will always score one point by dogfighting with Shion, and the opponent will be stunned for T seconds after dogfighting.
Since Asuka is slower than Shion, she decides to fight with Shion for
only one time during the match. It is also assumed that if Asuka and
Shion touch the buoy in the same time, the point will be given to Asuka
and Asuka could also fight with Shion at the buoy. We assume that in
such scenario, the dogfighting must happen after the buoy is touched by
Asuka or Shion.

The speed of Asuka is V1 m/s. The speed of Shion is V2 m/s. Is there any possibility for Asuka to win the match (to have higher score)?

Input

输入共2行。
第1行包含1个正整数n表示n个人。
第2行包含n个用空格隔开的正整数T1,T2,……,Tn其中第i个整数Ti示编号为i
的同学的信息传递对象是编号为Ti的同学，Ti≤n且Ti≠i
数据保证游戏一定会结束。

Output

输出共 1 行，包含 1 个整数，表示游戏一共可以进行多少轮。

Sample Input

5
2 4 2 3 1

Sample Output

3

HINT

题意

题解：

tarjan找SCC，然后找到最小的SCC就好了

两次 dfs也可以啦~

但是TM的会爆栈……

代码

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<stdio.h>
#include<vector>
#include<cstring>
#include<stack>
using namespace std;
#define maxn 300005

vector<int> G[maxn];
int pre[maxn],lowlink[maxn],sccno[maxn],dfs_clock,scc_cnt,num[maxn];
stack<int> S;
void dfs(int u)
{
    pre[u]=lowlink[u]=++dfs_clock;
    S.push(u);
    for(int i=;i<G[u].size();i++)
    {
        int v = G[u][i];
        if(!pre[v])
        {
            dfs(v);
            lowlink[u]=min(lowlink[u],lowlink[v]);
        }
        else if(!sccno[v])
        {
            lowlink[u]=min(lowlink[u],pre[v]);
        }
    }
    if(lowlink[u]==pre[u])
    {
        scc_cnt++;
        for(;;)
        {
            int x = S.top();S.pop();
            sccno[x]=scc_cnt;
            num[scc_cnt]++;
            if(x==u)break;
        }
    }
}
void find_scc(int n)
{
    dfs_clock = scc_cnt = ;
    memset(sccno,,sizeof(sccno));
    memset(pre,,sizeof(pre));
    for(int i=;i<n;i++)
        if(!pre[i])dfs(i);
}
int main()
{
    int n;
    scanf("%d",&n);
    int flag = ;
    for(int i=;i<n;i++)
    {
        int x;scanf("%d",&x);
        x--;
        if(i==x)
            flag = ;
        G[i].push_back(x);
    }
    if(flag)
    {
        printf("1\n");
        return ;
    }
    find_scc(n);
    int ans = ;
    for(int i=;i<scc_cnt;i++)
        if(num[i]>)
            ans = min(ans,num[i]);
    printf("%d\n",ans);
}