【bzoj1589】[Usaco2008 Dec]Trick or Treat on the Farm 采集糖果

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.

题解
tarjan缩点

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<set>
#include<ctime>
#include<vector>
#include<cmath>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#define mod 1000000007
#define ll long long 
#define N 25
#define inf 0x7fffffff
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int tot,ret[100005],Next[100005],Head[100005],to[100005];
int dfn[100005],low[100005],q[100005],tim;
bool inq[100005];
int num[100005],ans[100005],bl[100005],scc,n,top;
void tarjan(int u)
{
    dfn[u]=low[u]=++tim;
    q[++top]=u;inq[u]=1;
    if (inq[to[u]]) low[u]=min(low[u],low[to[u]]);
    else if (!dfn[to[u]]){tarjan(to[u]);low[u]=min(low[u],low[to[u]]);}
    if (low[u]==dfn[u])
    {
        scc++;int now=0;
        while (now!=u)
        {
            now=q[top--];
            bl[now]=scc;
            num[scc]++;
            inq[now]=0;
        }
    }
}
inline void ins(int u,int v)
{
    ret[++tot]=v;Next[tot]=Head[u];Head[u]=tot;
}
void build()
{
    for (int i=1;i<=n;i++)
        if (bl[i]!=bl[to[i]]) ins(bl[i],bl[to[i]]);
}
int solve(int u)
{
    if (ans[u]) return ans[u];
    ans[u]+=num[u];
    for (int i=Head[u];i;i=Next[i])
        ans[u]+=solve(ret[i]);
    return ans[u];
}
int main()
{
    n=read();
    for (int i=1;i<=n;i++)
    {
        int x=read();
        to[i]=x;
    }
    for (int i=1;i<=n;i++)
        if (!dfn[i]) tarjan(i);
    build();
    for (int i=1;i<=n;i++) printf("%d\n",solve(bl[i]));
    return 0;
}