QOJ.ac
QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#56714 | #238. Distinct Values | mango8023 | WA | 352ms | 9276kb | C++ | 1013b | 2022-10-21 09:14:37 | 2022-10-21 09:14:39 |
Judging History
answer
#include <cstdio>
#include <iostream>
#include <set>
using namespace std;
const int N = 1E5+5;
int n, k, pre[N], ans[N];
int main(){
int t = 0;
scanf("%d", &t);
while(t--){
scanf("%d%d", &k, &n);
for(int i=1;i<=k;i++) pre[i] = i;
for(int i=0;i<n;i++){
int l, r;
scanf("%d%d", &l, &r);
pre[r] = min(pre[r], l);
}
for(int i=1;i<k;i++){
pre[i] = min(pre[i], pre[i+1]);
}
int num = 1;
set<int> val;
for(int i=1;i<=k;i++) val.insert(i);
for(int i=1;i<=k;i++){ // O(k*log(k) + k + k*(1+log(k)))
while(num < pre[i]){
val.insert(ans[num]); // O(k log k)
num++; // O(k)
}
ans[i] = *val.begin(); // O(1)
val.erase(ans[i]); // O(log k)
}
for(int i=1;i<=k;i++){
printf("%d%c", ans[i], i==k?'\n':' ');
}
}
return 0;
}
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 0
Wrong Answer
time: 352ms
memory: 9276kb
input:
11116 10 2 5 5 5 6 10 1 7 10 10 1 2 6 10 1 2 5 10 1 6 7 10 2 8 9 7 10 10 2 1 4 6 10 10 4 8 8 10 10 3 6 1 5 10 3 8 8 10 10 8 10 10 4 6 10 1 5 2 6 1 2 10 3 4 4 4 8 4 8 10 4 1 5 1 2 5 5 2 4 10 4 2 5 9 10 6 7 2 4 10 1 5 6 10 4 10 10 8 10 2 5 10 10 10 1 1 2 10 4 7 8 5 6 7 9 10 10 10 4 3 7 6 6 8 10 3 4 10...
output:
1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 2 3 1 1 1 1 2 3 1 1 1 1 1 1 1 2 3 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 2 3 1 1 2 3 1 1 1 1 2 3 1 1 1 2 3 4 1 1 1 1 1 1 1 1 1 1 1 1 2 3 1 2 1 2 3 4 1 1 2 3 1 1 1 1 1 1 2 3 1 1 1 2 1 3 4 1 1 1 1 1 1 1 2 3 4 1 2 1 1 2 1 1 1 1 1 2 1 1 1 1 1 1 1 2 3 1 1 1 2 3 ...
result:
wrong answer 2nd lines differ - expected: '1 1 1 1 1 1 1 2 3 4', found: '1 1 1 1 1 1 1 1 2 3'