强连通分量

USACO08DEC

题目描述

Every year in Wisconsin the cows celebrate the USA autumn holiday of Halloween by dressing up in costumes and collecting candy that Farmer John leaves in the N (1 <= N <= 100,000) stalls conveniently numbered 1..N.

Because the barn is not so large, FJ makes sure the cows extend their fun by specifying a traversal route the cows must follow. To implement this scheme for traveling back and forth through the barn, FJ has posted a 'next stall number' next_i (1 <= next_i <= N) on stall i that tells the cows which stall to visit next; the cows thus might travel the length of the barn many times in order to collect their candy.

FJ mandates that cow i should start collecting candy at stall i. A cow stops her candy collection if she arrives back at any stall she has already visited.

Calculate the number of unique stalls each cow visits before being forced to stop her candy collection.

POINTS: 100

输入格式

* Line 1: A single integer: N

* Lines 2..N+1: Line i+1 contains a single integer: next_i

输出格式

* Lines 1..N: Line i contains a single integer that is the total number of unique stalls visited by cow i before she returns to a stall she has previously visited.

题意翻译

题目描述

每年，在威斯康星州，奶牛们都会穿上衣服，收集农夫约翰在N(1<=N<=100，000)个牛棚隔间中留下的糖果，以此来庆祝美国秋天的万圣节。

由于牛棚不太大，FJ通过指定奶牛必须遵循的穿越路线来确保奶牛的乐趣。为了实现这个让奶牛在牛棚里来回穿梭的方案，FJ在第i号隔间上张贴了一个“下一个隔间”Next_i(1<=Next_i<=N)，告诉奶牛要去的下一个隔间；这样，为了收集它们的糖果，奶牛就会在牛棚里来回穿梭了。

FJ命令奶牛i应该从i号隔间开始收集糖果。如果一只奶牛回到某一个她已经去过的隔间，她就会停止收集糖果。

在被迫停止收集糖果之前，计算一下每头奶牛要前往的隔间数（包含起点）。

输入格式

第1行 整数n。

第2行到n+1行 每行包含一个整数 next_i 。

输出格式

n行，第i行包含一个整数，表示第i只奶牛要前往的隔间数。

样例解释

有4个隔间

隔间1要求牛到隔间1

隔间2要求牛到隔间3

隔间3要求牛到隔间2

隔间4要求牛到隔间3

牛1，从1号隔间出发，总共访问1个隔间；

牛2，从2号隔间出发，然后到三号隔间，然后到2号隔间，终止，总共访问2个隔间；

牛3，从3号隔间出发，然后到2号隔间，然后到3号隔间，终止，总共访问2个隔间；

牛4，从4号隔间出发，然后到3号隔间，然后到2号隔间，然后到3号隔间，终止，总共访问3个隔间。

翻译提供者：吃葡萄吐糖

输入输出样例

输入 #1复制

4 
1 
3 
2 
3 

输出 #1复制

1 
2 
2 
3 

说明/提示

Four stalls.

* Stall 1 directs the cow back to stall 1.

* Stall 2 directs the cow to stall 3

* Stall 3 directs the cow to stall 2

* Stall 4 directs the cow to stall 3

Cow 1: Start at 1, next is 1. Total stalls visited: 1.

Cow 2: Start at 2, next is 3, next is 2. Total stalls visited: 2. Cow 3: Start at 3, next is 2, next is 3. Total stalls visited: 2. Cow 4: Start at 4, next is 3, next is 2, next is 3. Total stalls visited: 3.

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int nexts[N],ans[N];
vector<int>G[N];
int n;
int sum,color[N],low[N],ins[N],dfn[N],col;
int Lemon[N];
stack<int>s;
void tarjan(int x)
{
    sum++;
    dfn[x]=low[x]=sum;
    s.push(x);
    ins[x]=1;
    for(int i=0; i<G[x].size(); i++)
    {
        if(ins[G[x][i]]==0)
        {
            tarjan(G[x][i]);
            low[x]=min(low[x],low[G[x][i]]);
        }
        else if(ins[G[x][i]]==1)
            low[x]=min(low[x],dfn[G[x][i]]);
    }
    if(dfn[x]==low[x])
    {
        col++;
        int node;
        do
        {
            node=s.top();
            s.pop();
            color[node]=col;
            ins[node]=-1;
        }
        while(node!=x);
    }
    return ;
}
void search(int root,int x,int step)
{
    if(ans[x]!=0)
    {
        ans[root]=ans[x]+step;
        return ;
    }
    else search(root,nexts[x],step+1);
}
int main()
{
    cin>>n;
    for(int i=1; i<=n; i++)
    {
        cin>>nexts[i];
        G[i].push_back(nexts[i]);
        if(nexts[i]==i)ans[i]=1;
    }
    for(int i=1; i<=n; i++)
    {
        if(ins[i]==0)
            tarjan(i);
    }
    for(int i=1; i<=n; i++)
    {
        Lemon[color[i]]++;//记录环的大小
    }
    for(int i=1; i<=n; i++)
        if(Lemon[color[i]]!=1) ans[i]=Lemon[color[i]];//处理在环内的点
    for(int i=1; i<=n; i++)
        if(ans[i]==0) search(i,nexts[i],1);//处理在环外的点。
    for(int i=1; i<=n; i++)
        cout<<ans[i]<<endl;
    return 0;
}