BZOJ1589: [Usaco2008 Dec]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: 400  Solved: 220
[Submit][Status]

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

题解：

先tarjan，然后如果在一个>1的环内答案肯定就是这个环的大小，否则就是1+它的后继的ans

代码：

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<string>
#define inf 1000000000
#define maxn 500+100
#define maxm 500+100
#define eps 1e-10
#define ll long long
#define pa pair<int,int>
#define for0(i,n) for(int i=0;i<=(n);i++)
#define for1(i,n) for(int i=1;i<=(n);i++)
#define for2(i,x,y) for(int i=(x);i<=(y);i++)
#define for3(i,x,y) for(int i=(x);i>=(y);i--)
#define mod 1000000007
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=10*x+ch-'0';ch=getchar();}
    return x*f;
}
int n,ti,tot,top,cnt;
int go[100005],sta[100005],dfn[100005],low[100005];
int head[100005],s[100005],scc[100005],ans[100005];
struct data{int go,next;}e[100005];
inline void insert(int x,int y)
{
    e[++tot].go=y;e[tot].next=head[x];head[x]=tot;
}
inline void dfs(int x)
{
    dfn[x]=low[x]=++ti;sta[++top]=x;
    if(!dfn[go[x]]){dfs(go[x]);low[x]=min(low[x],low[go[x]]);}
    else if(!scc[go[x]])low[x]=min(low[x],dfn[go[x]]);
    if(low[x]==dfn[x])
    {
        cnt++;int now=0;
        while(now!=x)
        {
            now=sta[top--];
            scc[now]=cnt;
            s[cnt]++;
        }
    }
}
inline int solve(int x)
{
    if(ans[x])return ans[x];
    ans[x]=s[x];
    if(s[x]==1)ans[x]+=solve(e[head[x]].go);
    return ans[x];
}
int main()
{
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    n=read();
    for1(i,n)go[i]=read();
    for1(i,n)if(!dfn[i])dfs(i);
    for1(i,n)if(scc[i]!=scc[go[i]])insert(scc[i],scc[go[i]]);
    for1(i,n)printf("%d\n",solve(scc[i]));
    return 0;
}