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QOJ
| ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
|---|---|---|---|---|---|---|---|---|---|
| #258964 | #5455. TreeScript | andyc_03 | RE | 10ms | 6168kb | C++14 | 568b | 2023-11-20 14:47:14 | 2023-11-20 14:47:14 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
vector <int> G[maxn];
int n, f[maxn];
void dfs(int u)
{
f[u] = 0;
int mx = 0, mxx = 0;
for (int to:G[u]) {
dfs(to);
if (f[to] > mx) mxx = mx, mx = f[to];
else mxx = max(mxx, f[to]);
}
f[u] = max(mx, mxx + 1);
}
int main()
{
int T; scanf("%d", &T);
while (T--) {
scanf("%d", &n);
for (int i = 0; i <= n; i++) G[i].clear();
for (int i = 1; i <= n; i++) {
int x; scanf("%d", &x);
G[x].push_back(i);
}
dfs(1);
printf("%d\n", f[1]);
}
}
詳細信息
Test #1:
score: 100
Accepted
time: 1ms
memory: 6140kb
input:
2 3 0 1 2 7 0 1 2 2 1 4 1
output:
1 2
result:
ok 2 number(s): "1 2"
Test #2:
score: 0
Accepted
time: 10ms
memory: 6168kb
input:
1000 197 0 1 1 2 1 4 1 5 8 3 5 1 4 7 12 14 4 7 10 9 12 11 16 10 21 19 22 17 25 13 28 9 5 15 26 26 33 25 15 1 35 6 32 17 37 8 19 43 19 27 29 9 30 6 31 27 35 35 37 13 28 38 57 31 38 8 22 14 33 9 18 62 52 37 10 19 22 60 54 12 38 59 64 65 80 82 28 60 85 78 27 25 71 14 52 6 59 14 87 32 33 41 59 41 88 38 ...
output:
4 4 3 4 3 4 5 5 4 4 4 4 5 5 3 4 4 5 3 5 5 3 5 4 5 4 5 3 4 5 2 5 4 3 5 5 4 3 3 4 4 4 3 3 4 4 5 5 4 4 4 2 4 5 5 4 4 5 4 4 5 4 3 4 4 4 4 4 5 4 4 5 5 4 5 5 2 3 4 1 1 5 5 4 3 4 5 5 4 5 4 4 4 4 4 4 4 4 4 4 5 4 4 6 5 5 3 3 5 3 4 3 3 3 4 4 3 2 4 4 3 5 3 5 3 4 4 4 5 4 5 4 5 4 4 3 5 4 4 3 3 3 3 2 5 5 4 2 4 5 ...
result:
ok 1000 numbers
Test #3:
score: -100
Runtime Error
input:
1 200000 0 1 2 1 4 2 3 4 1 7 9 3 4 13 9 15 11 7 7 14 5 11 16 1 5 21 11 11 6 4 23 27 22 32 24 35 28 3 8 31 18 4 32 38 39 23 37 37 13 1 35 30 20 3 39 36 46 6 14 20 37 3 2 23 56 43 34 10 58 49 67 49 9 69 48 65 37 12 8 6 47 44 36 7 50 15 29 12 53 26 66 47 43 64 29 69 41 13 1 20 52 21 51 100 33 79 58 76 ...