Codeforces 920 反图联通块 线段树质因数暴力

A

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!
")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
ll Mod = 1000000007;
int num[205];
int pre[400];
int n;
int number;
bool check()
{
        for (int i = 1; i <= n; i++)
        {
                pre[i] = pre[i - 1] + pre[i];
                if (pre[i] < 1)
                {
                        return false;
                }
        }
        return true;
}
int main()
{
        int T;
        cin >> T;
        while (T--)
        {
                cin >> n >> number;
                for (int i = 1; i <= number; i++)
                {
                        cin >> num[i];
                }
                for (int i = 1; i <= 200; i++)
                {
                        for (int j = 1; j <= n; j++)
                        {
                                pre[j] = 0;
                        }
                        for (int j = 1; j <= number; j++)
                        {
                                if (i == 1)
                                {
                                        pre[num[j]] += 1;
                                        pre[num[j] + 1] += -1;
                                }
                                else
                                {
                                        pre[max(1, num[j] - i + 1)] += 1;
                                        pre[num[j] + i] += -1;
                                }
                        }
                        if (check())
                        {
                                cout << i << endl;
                                break;
                        }
                }
        }
        return 0;
}

B

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!
")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-8;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int  maxm = 300;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
ll mod = 3e7;
int le[5005];
int re[5005];
int wait[5005];
int main()
{
        int t;
        cin >> t;
        while (t--)
        {
                int n;
                scanf("%d", &n);
                for (int i = 0; i < n; i++)
                {
                        scanf("%d %d", &le[i], &re[i]);
                }
                int cur = le[0];
                for (int i = 0; i < n; i++)
                {
                        cur = max(cur, le[i]);
                        if (re[i] >= cur)
                        {
                                wait[i] = cur;
                                cur++;
                        }
                        else
                        {
                                wait[i] = 0;
                        }
                }
                for (int i = 0; i < n; i++)
                {
                        cout << wait[i];
                        if (i != n - 1)
                        {
                                cout << " ";
                        }
                }
                cout << endl;
        }
        return 0;
}

C

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!
")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
ll Mod = 1000000007;
int num[200005];
int pre[200005];
char f[200005];
int main()
{
        int n;
        cin >> n;
        for (int i = 1; i <= n; i++)
        {
                scanf("%d", num + i);
        }
        scanf("%s", f + 1);
        for (int i = 1; i < n; i++)
        {
                if (f[i] == '1')
                {
                        pre[i] = pre[i - 1] + 1;
                }
                else
                {
                        pre[i] = 0;
                }
        }
        for (int i = 1; i < n; i++)
        {
                if (pre[i] > pre[i + 1])
                {
                        sort(num + i - pre[i] + 1, num + i + 2);
                }
        }
        for (int i = 1; i < n; i++)
        {
                if (num[i] != i)
                {
                        cout << "NO" << endl;
                        return 0;
                }
        }
        cout << "YES" << endl;
        return 0;
}

E

 把所有点放在一个set里，每次取set中一个顶点，删去，遍历set，删去与此顶点邻接的顶点 

 因为遍历的过程中有两种结局1.删去某个结点 遍历成功 2.两点之间不存在边 遍历失败

 所以遍历的总复杂度为O(n+m) 再加上set的复杂度就是 O((n+m)log)

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!
")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
ll Mod = 1000000007;
set<int> need;
map<int, bool> mp[200005];
vector<int> ans;
int main()
{
        int n, m;
        int from, to;
        cin >> n;
        for (int i = 1; i <= n; i++)
        {
                need.insert(i);
        }
        cin >> m;
        for (int i = 1; i <= m; i++)
        {
                scanf("%d %d", &from, &to);
                mp[from][to] = mp[to][from] = 1;
        }
        while (!need.empty())
        {
                int todo = *need.begin();
                need.erase(todo);
                queue<int> que;
                ans.push_back(0);
                que.push(todo);
                while (!que.empty())
                {
                        queue<int> shan;
                        int cnt = que.front();
                        que.pop();
                        ans.back()++;
                        for (auto i : need)
                        {
                                if (!mp[cnt][i])
                                {
                                        que.push(i);
                                        shan.push(i);
                                }
                        }
                        while (!shan.empty())
                        {
                                need.erase(shan.front());
                                shan.pop();
                        }
                }
        }
        cout << ans.size() << endl;
        sort(ans.begin(), ans.end());
        for (auto i : ans)
        {
                cout << i << " ";
        }
        cout << endl;
        return 0;
}

F

如果知道到N的因数(N%i==0)数量级是N^(1/3)的这道题就很好做了 因为当N=1或者N=2时因数数目等于N 而1e6=2^20 每个数最多被修改7次

所以线段树维护一个最大值 一个sum值 当最大值不大于2时不用修改 大于二时递归下去暴力修改

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!
")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
ll Mod = 1000000007;
ll n, q;
int l, r;
ll dp[1000005];
struct node
{
        ll maxn, sum;
} tree[1200005];
void pushup(int x)
{
        tree[x].sum = tree[x << 1].sum + tree[x << 1 | 1].sum;
        tree[x].maxn = max(tree[x << 1].maxn, tree[x << 1 | 1].maxn);
}
void build(int x, ll value, int root = 1, int l = 1, int r = n)
{
        if (l > x || r < x)
        {
                return ;
        }
        if (l == x && r == x)
        {
                tree[root].sum = tree[root].maxn = value;
                return;
        }
        int mid = (l + r) >> 1;
        if (x <= mid)
        {
                build(x, value, root << 1, l, mid);
        }
        else
        {
                build(x, value, root << 1 | 1, mid + 1, r);
        }
        pushup(root);
}
void update(int xl, int xr, int root = 1, int l = 1, int r = n)
{
        if (l > r || l > xr || r < xl)
        {
                return;
        }
        if (xl <= l && xr >= r && tree[root].maxn <= 2)
        {
                return;
        }
        if (l == r)
        {
                tree[root].sum = tree[root].maxn = dp[tree[root].sum];
                return ;
        }
        int mid = (l + r) >> 1;
        if (xl <= mid)
        {
                update(xl, xr, root << 1, l, mid);
        }
        if (xr > mid)
        {
                update(xl, xr, root << 1 | 1, mid + 1, r);
        }
        pushup(root);
}
ll getsum(int xl, int xr, int root = 1, int l = 1, int r = n)
{
        if (l > r || l > xr || r < xl)
        {
                return 0;
        }
        if (xl <= l && xr >= r)
        {
                return tree[root].sum;
        }
        int mid = (l + r) >> 1;
        return getsum(xl, xr, root << 1, l, mid) + getsum(xl, xr, root << 1 | 1, mid + 1, r);
}
int main()
{
        cin >> n >> q;
        ll cnt;
        for (int i = 1; i <= 1000000; i++)
        {
                for (int j = i; j <= 1000000; j += i)
                {
                        dp[j]++;
                }
        }
        for (int i = 1; i <= n; i++)
        {
                scanf("%lld", &cnt);
                build(i, cnt);
        }
        for (int i = 1; i <= q; i++)
        {
                int now;
                ll value;
                int aim;
                cin >> now;
                if (now == 1)
                {
                        scanf("%d %d", &l, &r);
                        update(l, r);
                }
                else
                {
                        scanf("%d %d", &l, &r);
                        cout << getsum(l, r) << endl;
                }
        }
        return 0;
}