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=,f=;char ch=getchar();

     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

     while(ch>=''&&ch<=''){x=*x+ch-'';ch=getchar();}

     return x*f;

 }

 int n,ti,tot,top,cnt;

 int go[],sta[],dfn[],low[];

 int head[],s[],scc[],ans[];

 struct data{int go,next;}e[];

 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=;

         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]==)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 ;

 }