[Usaco2008 Dec][BZOJ1589] Trick or Treat on the Farm 采集糖果

1589: [Usaco2008 Dec]Trick or Treat on the Farm 采集糖果

Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 438 Solved: 244
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Description

每年万圣节，威斯康星的奶牛们都要打扮一番，出门在农场的N(1≤N≤100000)个牛棚里转悠，来采集糖果．她们每走到一个未曾经过的牛棚，就会采集这个棚里的1颗糖果．农场不大，所以约翰要想尽法子让奶牛们得到快乐．他给每一个牛棚设置了一个“后继牛棚”．牛棚i的后继牛棚是Xi．他告诉奶牛们，她们到了一个牛棚之后，只要再往后继牛棚走去，就可以搜集到很多糖果．事实上这是一种有点欺骗意味的手段，来节约他的糖果．第i只奶牛从牛棚i开始她的旅程．请你计算，每一只奶牛可以采集到多少糖果．

Input

第1行输入N，之后一行一个整数表示牛棚i的后继牛棚Xi，共N行．

Output

共N行，一行一个整数表示一只奶牛可以采集的糖果数量．

Sample Input

4 //有四个点
1 //1有一条边指向1
3 //2有一条边指向3
2 //3有一条边指向2
3

INPUT DETAILS:

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

Sample Output

1
2
2
3

HINT

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.

Source

Gold

tarjan缩环然后记忆化搜一下……（写完才反应过来一个点只有一个后继，根本不用数组模拟链表，开个数组记录一下就好了）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<vector>
#define MAX 100005
using namespace std;
int top,n,x,tot,sum,cnt,tot0;
int stack[MAX],inset[MAX],num[MAX],next[MAX],list[MAX],next0[MAX],list0[MAX],dfn[MAX],low[MAX],head[MAX],head0[MAX],ans[MAX],belong[MAX];
void insert(int x,int y)
{
    next[++tot]=head[x];
    head[x]=tot;
    list[tot]=y;
}
void insert0(int x,int y)
{
    next0[++tot0]=head0[x];
    head0[x]=tot0;
    list0[tot0]=y;
}
void dfs(int x)
{
    dfn[x]=low[x]=++sum;
    stack[++top]=x;
    inset[x]=1;
    for (int i=head[x];i;i=next[i])
    {
        if (!dfn[list[i]])
        {
            dfs(list[i]);
            low[x]=min(low[x],low[list[i]]);
        }
        if (inset[list[i]]) low[x]=min(low[x],dfn[list[i]]);
    }
    int y=-1;
    if (dfn[x]==low[x])
    {
        cnt++;
        while (y!=x)
        {
            y=stack[top--];
            inset[y]=0;
            belong[y]=cnt;
            num[cnt]++;
        }
    }
}
void tarjan()
{
    for (int i=1;i<=n;i++) if (!dfn[i]) dfs(i);
}
void rebuild()
{
    for (int i=1;i<=n;i++)
        for (int j=head[i];j;j=next[j])
            if (belong[i]!=belong[list[i]])
                insert0(belong[i],belong[list[j]]);
}
int search(int x)
{
    if (ans[x]!=-1) return ans[x];
    ans[x]=num[x];
    ans[x]+=search(list0[head0[x]]);
    return ans[x];
}
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
    {
        scanf("%d",&x);
        insert(i,x);
    }
    tarjan();
    rebuild();
    memset(ans,-1,sizeof(ans));
    for (int i=1;i<=n;i++) printf("%d
",search(belong[i]));
    return 0;
}