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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#864560 | #9678. 网友小 Z 的树 | KaXdd_# | 0 | 1ms | 16008kb | C++14 | 1.3kb | 2025-01-20 18:48:32 | 2025-01-20 18:48:34 |
Judging History
answer
#include"diameter.h"
#include<bits/stdc++.h>
//#define int long long
#define L(i, a, b) for (int i = (a); i <= (b); i++)
#define R(i, a, b) for (int i = (a); i >= (b); i--)
using namespace std;
pair<int, int> find_diameter(int subid, int n){
const int inf = 1e9;
if (n == 1) return {1, 1};
if (n == 2) return {1, 2};
if (n == 3){
if (in(1, 2, 3)) return {2, 3};
if (in(2, 1, 3)) return {1, 3};
return {1, 2};
}
int mx = 0, p = 0;
L(i, 1, n){
if ((i ^ 1) && (i ^ 2)){
int ret = query(1, 2, i);
if (ret > mx){
mx = ret;
p = i;
}
}
}
int x = p;
mx = p = 0;
L(i, 1, n){
if ((i ^ 2) && (i ^ x)){
int ret = query(2, x, i);
if (ret > mx){
mx = ret;
p = i;
}
}
}
int y = p;
mx = p = 0;
int xy = inf / 2;
L(i, 1, n){
if ((i ^ x) && (i ^ y)){
int ret = query(x, y, i);
xy = min(xy, ret);
if (ret > mx){
mx = ret;
p = i;
}
}
}
int xp = inf / 2, py = inf / 2;
L(i, 1, n){
if ((i ^ x) && (i ^ p)){
int ret = query(x, p, i);
xp = min(xp, ret);
}
}
L(i, 1, n){
if ((i ^ p) && (i ^ y)){
int ret = query(p, y, i);
py = min(py, ret);
}
}
if (xy == min({xy, xp, py})) return {x, y};
else if (xp = min({xy, xp, py})) return {x, p};
else return {p, y};
}
Details
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Subtask #1:
score: 0
Wrong Answer
Test #1:
score: 0
Wrong Answer
time: 1ms
memory: 16008kb
input:
1 100 25 1 3 2 18 3 8 4 18 5 14 6 22 7 18 8 10 9 11 10 12 11 25 12 16 13 11 14 17 15 17 16 25 17 2 18 20 19 18 20 12 21 1 22 17 23 14 24 1 50 1 37 2 27 3 10 4 25 5 16 6 17 7 10 8 36 9 16 10 6 11 48 12 2 13 28 14 30 15 10 16 44 17 31 18 1 19 6 20 7 21 30 22 42 23 45 24 23 25 27 26 39 27 45 28 48 29 4...
output:
WA
result:
wrong answer Wrong Answer
Subtask #2:
score: 0
Skipped
Dependency #1:
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Dependency #1:
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