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

2024-07-29 08:33:59   218人阅读

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+它的出边的ans

代码：

 1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<iostream> 7 #include<vector> 8 #include<map> 9 #include<set>10 #include<queue>11 #include<string>12 #define inf 100000000013 #define maxn 500+10014 #define maxm 500+10015 #define eps 1e-1016 #define ll long long17 #define pa pair<int,int>18 #define for0(i,n) for(int i=0;i<=(n);i++)19 #define for1(i,n) for(int i=1;i<=(n);i++)20 #define for2(i,x,y) for(int i=(x);i<=(y);i++)21 #define for3(i,x,y) for(int i=(x);i>=(y);i--)22 #define mod 100000000723 using namespace std;24 inline int read()25 {26     int x=0,f=1;char ch=getchar();27     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}28     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}29     return x*f;30 }31 int n,ti,tot,top,cnt;32 int go[100005],sta[100005],dfn[100005],low[100005];33 int head[100005],s[100005],scc[100005],ans[100005];34 struct data{int go,next;}e[100005];35 inline void insert(int x,int y)36 {37     e[++tot].go=y;e[tot].next=head[x];head[x]=tot;38 }39 inline void dfs(int x)40 {41     dfn[x]=low[x]=++ti;sta[++top]=x;42     if(!dfn[go[x]]){dfs(go[x]);low[x]=min(low[x],low[go[x]]);}43     else if(!scc[go[x]])low[x]=min(low[x],dfn[go[x]]);44     if(low[x]==dfn[x])45     {46         cnt++;int now=0;47         while(now!=x)48         {49             now=sta[top--];50             scc[now]=cnt;51             s[cnt]++;52         }53     }54 }55 inline int solve(int x)56 {57     if(ans[x])return ans[x];58     ans[x]=s[x];59     if(s[x]==1)ans[x]+=solve(e[head[x]].go);60     return ans[x];61 }62 int main()63 {64     freopen("input.txt","r",stdin);65     freopen("output.txt","w",stdout);66     n=read();67     for1(i,n)go[i]=read();68     for1(i,n)if(!dfn[i])dfs(i);69     for1(i,n)if(scc[i]!=scc[go[i]])insert(scc[i],scc[go[i]]);70     for1(i,n)printf("%d\n",solve(scc[i]));71     return 0;72 }
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BZOJ1589: [Usaco2008 Dec]Trick or Treat on the Farm 采集糖果


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