题目链接：
1 second
256 megabytes
standard input
standard output
Alyona decided to go on a diet and went to the forest to get some apples. There she unexpectedly found a magic rooted tree with root in the vertex 1, every vertex and every edge of which has a number written on.
The girl noticed that some of the tree's vertices are sad, so she decided to play with them. Let's call vertex v sad if there is a vertex u in subtree of vertex v such that dist(v, u) > au, where au is the number written on vertex u, dist(v, u) is the sum of the numbers written on the edges on the path from v to u.
Leaves of a tree are vertices connected to a single vertex by a single edge, but the root of a tree is a leaf if and only if the tree consists of a single vertex — root.
Thus Alyona decided to remove some of tree leaves until there will be no any sad vertex left in the tree. What is the minimum number of leaves Alyona needs to remove?
In the first line of the input integer n (1 ≤ n ≤ 105) is given — the number of vertices in the tree.
In the second line the sequence of n integers a1, a2, ..., an (1 ≤ ai ≤ 109) is given, where ai is the number written on vertex i.
The next n - 1 lines describe tree edges: ith of them consists of two integers pi and ci (1 ≤ pi ≤ n, - 109 ≤ ci ≤ 109), meaning that there is an edge connecting vertices i + 1 and pi with number ci written on it.
Print the only integer — the minimum number of leaves Alyona needs to remove such that there will be no any sad vertex left in the tree.
9
88 22 83 14 95 91 98 53 11
3 24
7 -8
1 67
1 64
9 65
5 12
6 -80
3 8
5 题意： 给一棵树,如果这个节点v与它的一个子节点v dist(u,v)>=a[u];那么这个节点就是sad节点,现在要你开始去掉叶子节点,问你最少去掉多少个节点才能使这棵树没有sad节点; 思路： dfs一发,dfs的时候一边找出这个节点到根的距离,一边维护这个节点到根节点路径上节点距离的最小值,因为dis[u]-min(dis[v])>a[u]就不满足了,这时dfs就可以return了;
如果满足就计算器加上这个点,继续dfs,最后的答案就是n-计数器的值; AC代码：
//#include <bits/stdc++.h>
#include <vector>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <cstring>
#include <algorithm>
#include <cstdio> using namespace std;
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<''||CH>'';F= CH=='-',CH=getchar());
for(num=;CH>=''&&CH<='';num=num*+CH-'',CH=getchar());
F && (num=-num);
}
int stk[], tp;
template<class T> inline void print(T p) {
if(!p) { puts(""); return; }
while(p) stk[++ tp] = p%, p/=;
while(tp) putchar(stk[tp--] + '');
putchar('\n');
} const LL mod=1e9+;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e5+;
const int maxn=; int n,cnt=,num=,head[N];
LL a[N],dis[N];
struct Edge
{
int fr,to,va,next;
}edge[*N];
void addedge(int s,int e,int val)
{
edge[cnt].to=e;
edge[cnt].va=val;
edge[cnt].next=head[s];
head[s]=cnt++;
}
void dfs(int now,int fa,LL mmin)
{
if(dis[now]-mmin<=a[now])num++;
else return;
for(int i=head[now];i!=-;i=edge[i].next)
{
int fr=edge[i].to;
if(fr!=fa)
{
dis[fr]=dis[now]+edge[i].va;
dfs(fr,now,min(mmin,dis[now]));
}
}
} int main()
{
mst(head,-);
read(n);
Riep(n)read(a[i]);
int p,c;
Riep(n-)
{
read(p);
read(c);
addedge(i+,p,c);
addedge(p,i+,c);
}
dis[]=;
dfs(,-,inf);
printf("%d\n",n-num);
return ;
}