场次
CF题解
ABC比较水，然而细节比较多，共耗时1h
D一个神秘的交互题，二分卡我1h结果输出答案的时候查询了一次炸了（警示后人）
E构造题上个厕所5min瞪出结论，10min写完
补F，赛时在想二分答案结果场后qzr嘴构造左0右1所以答案至多为2
实际上就是考虑什么时候取0和1
- 0比较简单，就是所有区间的公共点取0
- 1的话考虑
- 若某个点不为任何区间的右端点，则这个点填1，下个点填0
- 若某个点不为任何区间左端点，则这个点填1，前面的点填0
然后1其他随便填就行，除了有点难调没别的
// By wnn
#include<bits/stdc++.h>
// simple name
#define pii pair<int,int>
#define pll pair<long long, long long>
#define pque priority_queue
#define x1 x_1
#define y1 y_1
#define fir first
#define sec second
#define pb push_back
#define myfreopen freopen(".in", "r", stdin),freopen(".out", "w", stdout)
// function
#define ls(x) (x << 1)
#define rs(x) ((x << 1) | 1)
#define mid(l, r) ((l + r) >> 1)
#define debug(x) cerr << x << endl
#define dist(x, y, x2, y2) sqrt((x - x2) * (x - x2) + (y - y2) * (y - y2))
#define WA cerr << "Wrong Answer" << endl
#define init_inf32(x) memset(x, 0x3f, sizeof(x))
#define init_inf64(x) memset(x, 0x3fll, sizeof(x))
#define init_0(x) memset(x, 0, sizeof(x))
#define Dec(x) fixed << setprecision(x)
// val
#define eps 1e-9
#define inf32 0x3f3f3f3f
#define inf64 0x3f3f3f3f3f3f3f3fll
#define mod1 (int)(1e9 + 7)
#define mod2 998244353
#define PI acos(-1.0)
// god
#define int long long
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
// Def important function
inline void init();
inline void ever_init();
inline void solve();
// Init rnd()
mt19937 rnd(time(0) ^ clock());
// Constants
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, 1, -1};
const int N = 3005;int n, m;
int l[N], r[N], d[N], st[N], ed[N], ans[N], pos = -1;
inline void init(){}
inline void ever_init(){cin >> n >> m; pos = -1;init_0(d);init_0(st);init_0(ed);for(int i = 1; i <= m; i++) cin >> l[i] >> r[i];for(int i = 1; i <= m; i++){st[l[i]] = 1;ed[r[i]] = 1;d[r[i] + 1] -= 1;d[l[i]] += 1;}for(int i = 1; i <= n; i++){d[i] += d[i - 1];if(d[i] == m){pos = i;break;}}memset(ans, -1, sizeof(ans));if(pos == -1){for(int i = 1; i < n; i++){if(!ed[i]){pos = 1; ans[i] = 0; ans[i + 1] = 1; break;}if(!st[i + 1]){pos = 1; ans[i] = 1; ans[i + 1] = 0; break;}}if(pos == -1){
// WA;cout << 0 << " ";for(int i = 2; i <= n - 1; i++) cout << i << " ";cout << 1 << "\n";return ;}else{
// WA;int p = 2;for(int i = 1; i <= n; i++){if(ans[i] != -1) cout << ans[i] << " ";else cout << p++ << " ";}cout << "\n";}return ;}int x = 0;for(int i = 1; i <= n; i++)if(i != pos) cout << i + x << " ";else{cout << 0 << " ";x = -1;}cout << "\n";
}
inline void solve(){}signed main(){
// myfreopen;ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);init();int T = 1;cin >> T;while(T--){ever_init();solve();}return 0;
}
/*things to check:
* Will it MLE?
* Is array big enough?
* Do you need long long?
* Is inf big enough?
* max or min?
* Yes,No or YES,NO?
* Is there anything extra to output?
* Did you Countershoot?
* Have you measured the limit data?
* More measurements should be cleared!!!
*/
G比较神奇，想到往n * (n + 1) / 2上凑就可以了
H目前不会
)
