【JZOJ5221】A

description

---

analysis

• 想象一下，如果一棵子树内权值的最大值减最小值加一等于子树大小，那么权值就是连续的

• 没了

---

code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 200005
#define INF 1000000007
#define reg register int
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)
#define rep(i,a) for (reg i=last[a];i;i=next[i])
#define O3 __attribute__((optimize("-O3")))

using namespace std;

int last[MAXN],next[MAXN],tov[MAXN];
int a[MAXN],mx[MAXN],mn[MAXN],sum[MAXN];
int n,tot,ans;

O3 inline int read()
{
	int x=0,f=1;char ch=getchar();
	while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
	while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
O3 inline void link(int x,int y)
{
	next[++tot]=last[x],last[x]=tot,tov[tot]=y;
}
O3 inline void dfs(int x)
{
	sum[x]=1,mx[x]=x,mn[x]=x;
	rep(i,x)dfs(tov[i]),sum[x]+=sum[tov[i]],mx[x]=max(mx[x],mx[tov[i]]),mn[x]=min(mn[x],mn[tov[i]]);
	if (mx[x]-mn[x]+1==sum[x])ans++;
}
O3 int main()
{
	//freopen("T2.in","r",stdin);
	n=read();
	fo(i,2,n)
	{
		int x=read(),y=read();
		link(x,y),++a[y];
	}
	fo(i,1,n)if (!a[i])
	{
		dfs(i);
		break;
	}
	printf("%d\n",ans);
	return 0;
}