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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 |
---|---|---|---|---|---|---|---|---|---|
#139753 | #238. Distinct Values | qwerty | 100 ✓ | 245ms | 9268kb | C++20 | 1.2kb | 2023-08-14 14:39:00 | 2023-08-14 14:39:59 |
Judging History
answer
#include <bits/stdc++.h>
#define maxn 100005
using namespace std;
int nxt[maxn], ans[maxn];
int n, m;
void solve() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) {
nxt[i] = 0;
}
for (int i = 1; i <= m; i++) {
int l, r;
scanf("%d%d", &l, &r);
nxt[l] = max(nxt[l], r);
}
for (int i = 1; i <= n; i++) {
nxt[i] = max(nxt[i - 1], nxt[i]);
if (nxt[i] < i)
nxt[i] = i;
}
set<int> st;
for (int i = 1; i <= n; i++) {
st.insert(i);
}
for (int i = 1; i <= n; i++) {
int now = *st.begin();
st.erase(now);
ans[now] = i;
for (;;) {
now = nxt[now] + 1;
set<int> ::iterator it = st.lower_bound(now) ;
if (it == st.end())
break;
else {
now = *it;
st.erase(it);
}
ans[now] = i;
}
}
for (int i = 1; i <= n; i++) {
printf("%d ", ans[i]);
}
putchar('\n');
}
int main() {
int T;
scanf("%d", &T);
while (T--) {
solve();
}
}
詳細信息
Test #1:
score: 100
Accepted
time: 245ms
memory: 9268kb
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 2 3 4 1 1 2 3 4 5 1 1 1 1 1 1 2 3 4 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 2 3 4 1 2 3 4 1 1 2 3 4 5 1 2 3 4 5 1 1 1 1 1 1 1 1 1 1 1 1 1 2 3 1 2 3 4 5 1 2 3 4 5 1 1 1 1 2 3 4 5 1 1 1 2 3 4 5 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 2 ...
result:
ok 11116 lines