BZOJ 4756.「USACO 2017 · January」Promotion Counting

屑题,一眼秒。
听说可以离散化然后树状数组容斥掉兄弟的贡献?

如果把 DFS 序搞出来,就是一个二维偏序了。
但是我还是喜欢写线段树合并!
多优美啊!

代码:

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#include <cstdio>
using namespace std;
const int N = 1e5;
const int C = 1e9;
int n,a[N + 5];
int to[N + 5],pre[N + 5],first[N + 5];
inline void add(int u,int v)
{
static int tot = 0;
to[++tot] = v,pre[tot] = first[u],first[u] = tot;
}
struct node
{
int sum;
int ls,rs;
} seg[(N << 6) + 10];
int rt[N + 5];
void insert(int x,int &p,int tl,int tr)
{
static int tot = 0;
if(!p)
p = ++tot;
++seg[p].sum;
if(tl == tr)
return ;
int mid = tl + tr >> 1;
x <= mid ? insert(x,seg[p].ls,tl,mid) : insert(x,seg[p].rs,mid + 1,tr);
}
int query(int l,int r,int p,int tl,int tr)
{
if(!p || (l <= tl && tr <= r))
return seg[p].sum;
int ret = 0;
int mid = tl + tr >> 1;
l <= mid && (ret += query(l,r,seg[p].ls,tl,mid));
r > mid && (ret += query(l,r,seg[p].rs,mid + 1,tr));
return ret;
}
int merge(int x,int y)
{
if(!x || !y)
return x | y;
seg[x].sum += seg[y].sum;
seg[x].ls = merge(seg[x].ls,seg[y].ls),seg[x].rs = merge(seg[x].rs,seg[y].rs);
return x;
}
int ans[N + 5];
void dfs(int p,int fa)
{
for(register int i = first[p];i;i = pre[i])
if(to[i] ^ fa)
dfs(to[i],p),rt[p] = merge(rt[p],rt[to[i]]);
ans[p] = query(a[p] + 1,C,rt[p],1,C),insert(a[p],rt[p],1,C);
}
int main()
{
scanf("%d",&n);
for(register int i = 1;i <= n;++i)
scanf("%d",a + i);
int u;
for(register int i = 2;i <= n;++i)
scanf("%d",&u),add(u,i);
dfs(1,0);
for(register int i = 1;i <= n;++i)
printf("%d\n",ans[i]);
}