原題傳送門
首先易知,某個點的權值大於S時,是-1情況
貪心策略,在樹上跑時,某個點有一堆兒子,肯定是和向上擴展能擴展得最遠的兒子結合成一條鏈
如果知道了每個點向上擴展最多能擴展到哪裏,那麼問題是多麼簡單啊!
只需要記錄每個點屬於的鏈的頂端,dfs的時候看看能不能挑出屬於鏈頂端深度最小的兒子,可以就把自己歸到那條鏈,否則自己新開一條鏈
至於如何得到每個點向上擴展最多能擴展到哪裏,這個可以用倍增預處理掉
依然要注意一下longlong
Code:
#include <bits/stdc++.h>
#define maxn 100010
#define int long long
using namespace std;
struct Edge{
int to, next;
}edge[maxn << 1];
int num, head[maxn], sum[maxn], d[maxn], fa[maxn][25], top[maxn], up[maxn], ans, n, L, S, w[maxn];
inline int read(){
int s = 0, w = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
return s * w;
}
void addedge(int x, int y) { edge[++num] = (Edge){ y, head[x] }; head[x] = num; }
void dfs(int u, int pre){
sum[u] = sum[pre] + w[u], d[u] = d[pre] + 1;
for (int i = 0; fa[u][i]; ++i) fa[u][i + 1] = fa[fa[u][i]][i];//倍增處理父親
top[u] = u;//初始化
int suml = L - 1;//自己也算鏈中一個點
for (int i = 20; i >= 0; --i){
int tmp = top[u];
if ((1 << i) > suml || !fa[tmp][i] || sum[u] - sum[fa[tmp][i]] + w[fa[tmp][i]] > S) continue;
//鏈中點個數超過L||根本沒有那個祖先||點權和超過S,這些情況都是不行的
suml -= (1 << i), top[u] = fa[tmp][i];//更新當前剩餘的L,更新鏈頂端
}
for (int i = head[u]; i; i = edge[i].next) dfs(edge[i].to, u);//往下dfs
}
void dfs(int u){
int tmp = -1;
for (int i = head[u]; i; i = edge[i].next){
int v = edge[i].to;
dfs(v);
if (up[v] != v && (tmp == -1 || d[up[v]] < d[tmp])) tmp = up[v];
//up[v]表示v所在鏈的頂端節點
//兒子所在鏈頂端不能是兒子自身(否則不能和我組成一條鏈),而且我要選擇深度最小的鏈頂端
}
if (tmp == -1) ++ans, up[u] = top[u]; else up[u] = tmp;
//無法選出,自己作爲鏈的底端,答案+1
}
signed main(){
n = read(), L = read(), S = read();
for (int i = 1; i <= n; ++i){
w[i] = read();
if (w[i] > S) return puts("-1"), 0;
}
for (int i = 2; i <= n; ++i){
fa[i][0] = read(); addedge(fa[i][0], i);
}
dfs(1, 0); dfs(1);
printf("%d\n", ans);
return 0;
}