今天复习了下差分约束，发现用的是SPFA，这个复杂度…害，明天有空整个板子吧，估计也不太用得上了

花了好长时间搞训练赛的题，明天比赛时间刚好把题解写了，明天再学学网络流好了

CF 1798A Showstopper

题目链接

小的换给a大的换给b就行

#include <bits/stdc++.h>using namespace std;#define int long long
#define INF 0x3f3f3f3ftypedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;const int N = 11;void solve()
{int n;cin >> n;vector<int> a(n), b(n);for (int i = 0; i < n; i ++ ) cin >> a[i];for (int i = 0; i < n; i ++ ) cin >> b[i];for (int i = 0; i < n; i ++ ){if (a[i] > b[i]) swap(a[i], b[i]);}bool flag = true;for (int i = 0; i < n - 1; i ++ ){if (a[n - 1] < a[i]) flag = false;if (b[n - 1] < b[i]) flag = false;if (!flag) break;}if (flag) cout << "YES\n";else cout << "NO\n";
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}

CF 1798B Three Sevens

题目链接

记录每个点的最迟抽奖时间，之后排序判断即可

#include <bits/stdc++.h>using namespace std;#define int long long
#define INF 0x3f3f3f3ftypedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;const int N = 11;void solve()
{int m;cin >> m;map<int, int> mp;for (int i = 1; i <= m; i ++ ){int n;cin >> n;for (int j = 0; j < n; j ++ ){int x; cin >> x;mp[x] = i;}}vector<PII> mes;for (auto t : mp){mes.push_back(make_pair(t.second, t.first));}sort(mes.begin(), mes.end());vector<int> ans;if (mes.size() < m){cout << -1 << '\n';return;}int cnt = 0;for (int i = 1; i <= m; i ++ ){if (cnt == mes.size()){cout << -1 << '\n';return;}if (mes[cnt].first == i){ans.push_back(mes[cnt].second);cnt ++ ;}else if (mes[cnt].first < i){i -- ;cnt ++ ;continue;}else{cout << "-1\n";return;}}for (auto t : ans) cout << t << ' ';cout << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}

CF 1798C Candy Store

题目链接

四题里唯一有点价值的一题

a是b的倍数，b是c的倍数（abc都是数组），如果a的gcd能整除c的lcm，那就存在b且b是二者相除的结果

#include <bits/stdc++.h>using namespace std;#define int long long
#define INF 0x3f3f3f3ftypedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;const int N = 11;
const int mod = 0;int lcm(int a, int b)
{return a * b / __gcd(a, b);
}void solve()
{int n;cin >> n;vector<int> a(n), b(n);for (int i = 0; i < n; i ++ ) cin >> a[i] >> b[i];int ans = 1;int tmp1 = 0, tmp2 = 1;for (int i = 0; i < n; i ++ ){if (__gcd(tmp1, a[i] * b[i]) % lcm(tmp2, b[i]) == 0){// ans ++ ;tmp1 = __gcd(tmp1, a[i] * b[i]);tmp2 = lcm(tmp2, b[i]);}else{ans ++ ;tmp1 = 0, tmp2 = 1;tmp1 = __gcd(tmp1, a[i] * b[i]);tmp2 = lcm(tmp2, b[i]);}}cout << ans << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}

CF 1798D Shocking Arrangement

题目链接

双指针，尽可能让区间和等于0

#include <bits/stdc++.h>using namespace std;#define int long long
#define INF 0x3f3f3f3ftypedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;const int N = 11;void solve()
{int n;cin >> n;vector<int> a(n);for (int i = 0; i < n; i ++ ) cin >> a[i];sort(a.begin(), a.end());int res = a[n - 1] - a[0];if (abs(a[0]) >= res || abs(a[n - 1]) >= res){cout << "NO\n";return;}int tmp = 0;int i = 0, j = n - 1;vector<int> ans;while (i <= j){if (tmp >= 0){tmp += a[i];ans.push_back(a[i]);i ++ ;}else{tmp += a[j];ans.push_back(a[j]);j -- ;}if (tmp >= res){cout << "NO\n";return;}}cout << "YES\n";for (auto t : ans) cout << t << ' ';cout << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}