BZOJ 1589 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 MBSubmit: 785 Solved: 447
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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
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
//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.
2
2
3
//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.
HINT
Source
tarjin缩完点后乱搞即可
推荐拓扑序
#include <bits/stdc++.h>
#define ll long long
#define inf 1e9+10
#define eps 1e-7
using namespace std;
inline int read(){
int x=0;int f=1;char ch=getchar();
while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
return x*f;
}
const int MAXN=2e5+10;
struct node{
int y,next;
}e[MAXN];
int linkk[MAXN],len=0,n,m,x[MAXN],y[MAXN],f[MAXN],q[MAXN],cnt[MAXN],dfs_clock,dfn[MAXN],sum[MAXN],low[MAXN],ine[MAXN],tot,vis[MAXN],stark[MAXN],top;
inline void insert(int xx,int yy){
e[++len].y=yy;e[len].next=linkk[xx];linkk[xx]=len;
}
inline void tarjin(int st){
dfn[st]=low[st]=++dfs_clock;
vis[st]=1;stark[++top]=st;
for(int i=linkk[st];i;i=e[i].next){
if(!dfn[e[i].y]){
tarjin(e[i].y);
low[st]=min(low[st],low[e[i].y]);
}
else if(vis[e[i].y]) low[st]=min(low[st],dfn[e[i].y]);
}
if(low[st]==dfn[st]){
int k;tot++;
do{
k=stark[top--];
vis[k]=0;
ine[k]=tot;
sum[tot]++;
}while(k!=st);
}
}
void rebuild(){
len=0;memset(linkk,0,sizeof(linkk));
for(int i=1;i<=n;i++){
if(ine[i]!=ine[y[i]]){
insert(ine[y[i]],ine[i]);
cnt[ine[i]]++;
}
}
}
void topsort(){
int head=0,tail=0;
for(int i=1;i<=tot;i++){
if(!cnt[i]) q[++tail]=i,f[i]=sum[i];
}
while(head<tail){
int tn=q[++head];
for(int i=linkk[tn];i;i=e[i].next){
cnt[e[i].y]--;f[e[i].y]=max(f[tn],f[e[i].y]);
if(!cnt[e[i].y]) q[++tail]=e[i].y,f[e[i].y]+=sum[e[i].y];
}
}
}
int main(){
n=read();
for(int i=1;i<=n;i++){
y[i]=read();
insert(i,y[i]);
}
for(int i=1;i<=n;i++){
if(!dfn[i]) tarjin(i);
}
rebuild();
topsort();
for(int i=1;i<=n;i++){
printf("%d\n",f[ine[i]]);
}
return 0;
}
/*
4
1
3
2
3
*/
