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 MBDescription
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
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.
解:强联通分量;
tarjan缩点;
then:记忆化搜索,sovle函数;
注意:first数组清零;(!!!)
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cstdio> #include<cstring> #include<algorithm> using std::min; const int N=100010; struct node { int from,to,next; }e[N]; int first[N],book[N]; int dfn[N],low[N],stack[N],cd[N],color[N],size[N]; int tot=0,top=0,num=0; int cnt=0; void insert(int v,int u) { e[++cnt].to=u;e[cnt].from=v;e[cnt].next=first[v];first[v]=cnt; } void tarjan(int x) { dfn[x]=low[x]=++tot; book[x]=1,stack[++top]=x; for(int i=first[x];i;i=e[i].next) { if(!dfn[e[i].to]) tarjan(e[i].to),low[x]=min(low[x],low[e[i].to]); else if(book[e[i].to]) low[x]=min(low[x],low[e[i].to]); } if(dfn[x]==low[x]) { ++num; while(stack[top]!=x) { book[stack[top]]=0; color[stack[top]]=num; size[num]++; top--; } book[x]=0;color[x]=num;size[num]++;top--; } } int ans[N]; int sovle(int x) { if(ans[x]) return ans[x]; ans[x]=size[x]; if(e[first[x]].to) ans[x]+=sovle(e[first[x]].to); return ans[x]; } int main() { int n; scanf("%d",&n); int x; for(int i=1;i<=n;i++) scanf("%d",&x),insert(i,x); for(int i=1;i<=n;i++) if(!dfn[i]) tarjan(i); cnt=0; memset(first,0,sizeof(first)); for(int i=1;i<=n;i++) { if(color[e[i].from]!=color[e[i].to]) cd[color[e[i].from]]++,insert(color[e[i].from],color[e[i].to]); } for(int i=1;i<=n;i++) printf("%d\n",sovle(color[i])); return 0; }
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