[Usaco08dec] Trick or Treat on the Farm

Title Description

Every Halloween, Wisconsin cows dress up and walk around the farm's N-barn to collect sweets. Each of them goes to a barn that has never passed and collects 1 candies from the shed.

The farm is not big, so John will try his best to make the cows happy. He set up a "successor barn" to every barn. The successor of the barn I was next_i he told the cows that after they had reached a barn, they had to go to the next barn, You can collect a lot of sweets. In fact, it is a kind of deceptive means to save his candy.

The first cow starts her journey from the barn I. Please calculate how many candies each cow can collect.

Input/output format

Input format:

• Line 1: A single integer:n

• Lines 2..n+1:line i+1 contains a single integer:next_i

Output format:

• Lines 1..n:line I contains a single integer that's the total number of unique stalls visited by cow I before she returns To a stall she had previously visited.

Input and Output Sample input example # #:

Sample # # of output:

Description

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 was 3, next is 2. Total stalls Visited:2 Cow 3:start at 3, Next was 2, next is 3. Total stalls visited:2 Cow 4:start at 4, Next was 3, Next was 2, next is 3. Total Stalls visited:3.

• The subject n<=100000, first consider the memory search, because the pure search in certain extreme situations will certainly be stuck (n^2).
• But how does the memory search come true?
• By observing the sample, we find that there is a ring.
• Tarjan indent + memory search. (the Great God said no Tarjan).
• The answer to all the points on the ring is the length of the ring, and the rest of the points? Because the other points are pointing to a ring, the memory search can be (DP).
• Expect to score 100 points.

1#include <cstdio>2#include <algorithm>3#include <iostream>4 #defineTime Dscada5 using namespacestd;6 7 intn,l,cnt,top,next[100050],summ[100050],pre[100050],time,last[100050],other[100050],f[100050],dfn[100050],low[100050],stack[100050],huan[100050];8 BOOLcir[100050],vis[100050];9 Ten voidAddintUintv) { Onepre[++l]=Last[u]; Alast[u]=l; -other[l]=v; - } the  - voidTarjan (intx) { -dfn[x]=low[x]=++Time ; -stack[++top]=x; +vis[x]=1; -      for(intP=LAST[X]; P p=Pre[p]) { +         intq=Other[p]; A         if(!Dfn[q]) { at Tarjan (q); -low[x]=min (low[x],low[q]); -}Else if(Vis[q]) low[x]=min (low[x],dfn[q]); -     } -     if(dfn[x]==Low[x]) { -         intsum=0; incnt++; -         intnow=stack[top--]; tocir[now]=1; +vis[now]=0; -huan[now]=CNT; thesum++; *          while(now!=x) { $now=stack[top--];Panax Notoginsengcir[now]=1; -vis[now]=0; thehuan[now]=CNT; +sum++; A         } thesumm[cnt]=sum; +     } - } $  $ intDfsintx) { -     if(F[x])returnF[x]; -     if(next[x]==x) { thef[x]=1; -         returnF[x];Wuyi     } the     if(Cir[x]) { -f[x]=Summ[huan[x]]; Wu         returnF[x]; -     } About     if(Cir[next[x]]) { $f[x]=summ[huan[next[x]]]+1; -         returnF[x]; -     } -F[x]=dfs (Next[x]) +1; A     returnF[x]; + } the  - intMain () { $scanf"%d",&n); the      for(intI=1; i<=n; i++) { the         intx; thescanf"%d",&x); thenext[i]=x; -         if(x!=i) Add (i,x);//There is no need to build edges here, use the next array.  in     } the      for(intI=1; i<=n; i++)  the         if(!Dfn[i]) Tarjan (i);//tarjan Reduction Point About      for(intI=1; i<=n; i++)if(summ[huan[i]]==1) cir[i]=0; the     //for (int i=1; i<=n; i++) printf ("%d\n", Cir[i]); the      for(intI=1; i<=n; i++) printf ("%d\n", DFS (i));//Memory Search the     return 0; +}

[Usaco08dec] Trick or Treat on the Farm