「BZOJ1589」[Usaco2008 Dec] Trick or Treat on the Farm 采集糖果
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
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.
题解
先缩个点然后乱搞即可。。
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#include<iostream> #include<cstdio> #include<cstring> #define ll long long using namespace std; inline ll read() { ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int n,tim,top,scc,cnt; int to[100005],q[100005],dfn[100005],low[100005]; int head[100005],hav[100005],belong[100005],ans[100005]; bool inq[100005]; struct data{int to,next;}e[100005]; void insert(int u,int v) {e[++cnt].to=v;e[cnt].next=head[u];head[u]=cnt;} void tarjan(int x) { int now=0; dfn[x]=low[x]=++tim; q[++top]=x;inq[x]=1; if(inq[to[x]])low[x]=min(low[x],dfn[to[x]]); else if(!dfn[to[x]]){tarjan(to[x]);low[x]=min(low[x],low[to[x]]);} if(low[x]==dfn[x]) { scc++; while(now!=x) { now=q[top--]; belong[now]=scc; hav[scc]++; inq[now]=0; } } } void rebuild() { for(int i=1;i<=n;i++) { if(belong[i]!=belong[to[i]]) insert(belong[i],belong[to[i]]); } } int solve(int x) { if(ans[x])return ans[x]; ans[x]=hav[x]; if(hav[x]==1) ans[x]+=solve(e[head[x]].to); return ans[x]; } int main() { n=read(); for(int i=1;i<=n;i++) to[i]=read(); for(int i=1;i<=n;i++) if(!dfn[i])tarjan(i); rebuild(); for(int i=1;i<=n;i++) printf("%d\n",solve(belong[i])); return 0; } |
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