• [POJ2761]Feed the dogs


    Problem

    查询区间第k大,但保证区间不互相包含(可以相交)

    Solution

    只需要对每个区间左端点进行排序,那它们的右端点必定单调递增,不然会出现区间包含的情况。
    所以我们暴力对下一个区间加上这个区间没有的点,删去下个区间没有的点。
    因为每个点最多被加入,删除1次,所以时间复杂度为O(nlogn)

    Notice

    当相邻两段区间不相交时,那么我们要先加入点,在删去点。

    Code

    非旋转Treap

    #include<cmath>
    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    #define sqz main
    #define ll long long
    #define reg register int
    #define rep(i, a, b) for (reg i = a; i <= b; i++)
    #define per(i, a, b) for (reg i = a; i >= b; i--)
    #define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
    const int INF = 1e9, N = 100000, M = 50000;
    const double eps = 1e-6, phi = acos(-1.0);
    ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
    ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
    if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
    void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
    int point = 0, T[N + 5], root, ans[M + 5];
    struct Node
    {
        int left, right, ask, id;
    }Q[M + 5];
    struct node
    {
        int Size[N + 5], Val[N + 5], Level[N + 5], Son[2][N + 5];
        inline void up(int u)
        {
            Size[u] = Size[Son[0][u]] + Size[Son[1][u]] + 1;
        }
        int Newnode(int v)
        {
            int u = ++point;
            Val[u] = v, Level[u] = rand();
            Size[u] = 1, Son[0][u] = Son[1][u] = 0;
            return u;
        }
        int Merge(int X, int Y)
        {
            if (X * Y == 0) return X + Y;
            if (Level[X] < Level[Y])
            {
                Son[1][X] = Merge(Son[1][X], Y);
                up(X); return X;
            }
            else
            {
                Son[0][Y] = Merge(X, Son[0][Y]);
                up(Y); return Y;
            }
        }
        void Split(int u, int t, int &x, int &y)
        {
            if (!u)
            {
                x = y = 0;
                return;
            }
            if (Val[u] <= t) x = u, Split(Son[1][u], t, Son[1][u], y);
            else y = u, Split(Son[0][u], t, x, Son[0][u]);
            up(u);
        }
        void Build(int l, int r)
        {
            int last, s[N + 5], top = 0;
            rep(i, l, r)
            {
                int u = Newnode(T[i]);
                last = 0;
                while (top && Level[s[top]] > Level[u])
                {
                    up(s[top]);
                    last = s[top--];
                }
                if (top) Son[1][s[top]] = u;
                Son[0][u] = last;
                s[++top] = u;
            }
            while (top) up(s[top--]);
            root = s[1];
        }
    
        int Find_num(int u, int t)
        {
        	if (t <= Size[Son[0][u]]) return Find_num(Son[0][u], t);
        	else if (t <= Size[Son[0][u]] + 1) return u;
        	else return Find_num(Son[1][u], t - Size[Son[0][u]] - 1);
    	}
        void Insert(int v)
        {
            int t = Newnode(v), x, y;
            Split(root, v, x, y);
            root = Merge(Merge(x, t), y);
        }
        void Delete(int v)
        {
            int x, y, z;
            Split(root, v, x, z), Split(x, v - 1, x, y);
            root = Merge(Merge(x, Merge(Son[0][y], Son[1][y])), z);
        }
    }Treap;
    int cmp(Node X, Node Y)
    {
    	return X.left < Y.left || (X.left == Y.left && X.right < Y.right);
    }
    int sqz()
    {
    	int n = read(), m = read();
    	rep(i, 1, n) T[i] = read();
    	rep(i, 1, m) Q[i].left = read(), Q[i].right = read(), Q[i].ask = read(), Q[i].id = i;
    	sort(Q + 1, Q + m + 1, cmp);
        rep(i, Q[1].left, Q[1].right) Treap.Insert(T[i]);
        ans[Q[1].id] = Treap.Val[Treap.Find_num(root, Q[1].ask)];
        rep(i, 2, m)
        {
            rep(j, Q[i - 1].right + 1, Q[i].right) Treap.Insert(T[j]);
            rep(j, Q[i - 1].left, Q[i].left - 1) Treap.Delete(T[j]);
            ans[Q[i].id] = Treap.Val[Treap.Find_num(root, Q[i].ask)];
        }
        rep(i, 1, m) printf("%d
    ", ans[i]);
        return 0;
    }
    

    旋转Treap

    #include<cmath>
    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    #define sqz main
    #define ll long long
    #define reg register int
    #define rep(i, a, b) for (reg i = a; i <= b; i++)
    #define per(i, a, b) for (reg i = a; i >= b; i--)
    #define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
    const int INF = 1e9, N = 100000, M = 50000;
    const double eps = 1e-6, phi = acos(-1.0);
    ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
    ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
    if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
    void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
    int point = 0, T[N + 5], root, ans[M + 5];
    struct node
    {
        int left, right, ask, id;
    }Q[M + 5];
    int cmp(node X, node Y)
    {
    	return X.left < Y.left || (X.left == Y.left && X.right < Y.right);
    }
    struct Node
    {
        int Val[N + 5], Son[2][N + 5], Level[N + 5], Size[N + 5], Num[N + 5];
        inline void up(int u)
        {
            Size[u] = Size[Son[0][u]] + Size[Son[1][u]] + Num[u];
        }
        inline void Lturn(int &x)
        {
            int y = Son[1][x]; Son[1][x] = Son[0][y]; Son[0][y] = x;
            up(x), up(y); x = y;
        }
        inline void Rturn(int &x)
        {
            int y = Son[0][x]; Son[0][x] = Son[1][y]; Son[1][y] = x;
            up(x), up(y); x = y;
        }
        inline void Newnode(int &u, int v)
        {
            u = ++point;
            Level[u] = rand(), Val[u] = v;
            Num[u] = Size[u] = 1, Son[0][u] = Son[1][u] = 0;
        }
    
        void Insert(int &u, int t)
        {
            if (!u)
            {
                Newnode(u, t);
                return;
            }
            Size[u]++;
            if (t == Val[u]) Num[u]++;
            else if (t < Val[u])
            {
                Insert(Son[0][u], t);
                if (Level[Son[0][u]] < Level[u]) Rturn(u);
            }
            else
            {
                Insert(Son[1][u], t);
                if (Level[Son[1][u]] < Level[u]) Lturn(u);
            }
        }
        void Delete(int &u, int t)
        {
            if (!u) return;
            if (Val[u] == t)
            {
                if (Num[u] > 1)
                {
                    Size[u]--, Num[u]--;
                    return;
                }
                if (Son[0][u] * Son[1][u] == 0) u = Son[0][u] + Son[1][u];
                else if (Level[Son[0][u]] < Level[Son[1][u]]) Rturn(u), Delete(u, t);
                else Lturn(u), Delete(u, t);
            }
            else if (t < Val[u]) Size[u]--, Delete(Son[0][u], t);
            else Size[u]--, Delete(Son[1][u], t);
        }
    
        int Find_num(int u, int t)
        {
            if (!u) return 0;
            if (t <= Size[Son[0][u]]) return Find_num(Son[0][u], t);
            else if (t <= Size[Son[0][u]] + Num[u]) return u;
            else return Find_num(Son[1][u], t - Size[Son[0][u]] - Num[u]);
        }
    }Treap;
    int sqz()
    {
        int n = read(), m = read();
        rep(i, 1, n) T[i] = read();
        rep(i, 1, m) Q[i].left = read(), Q[i].right = read(), Q[i].ask = read(), Q[i].id = i;
        sort(Q + 1, Q + m + 1, cmp);
        rep(i, Q[1].left, Q[1].right) Treap.Insert(root, T[i]);
        ans[Q[1].id] = Treap.Val[Treap.Find_num(root, Q[1].ask)];
        rep(i, 2, m)
        {
            if (Q[i].left <= Q[i - 1].right)
            {
                rep(j, Q[i - 1].left, Q[i].left - 1) Treap.Delete(root, T[j]);
                rep(j, Q[i - 1].right + 1, Q[i].right) Treap.Insert(root, T[j]);
            }
            else
            {
                rep(j, Q[i - 1].left, Q[i - 1].right) Treap.Delete(root, T[j]);
                rep(j, Q[i].left, Q[i].right) Treap.Insert(root, T[j]);
            }
            ans[Q[i].id] = Treap.Val[Treap.Find_num(root, Q[i].ask)];
        }
        rep(i, 1, m) printf("%d
    ", ans[i]);
        return 0;
    }
    
  • 相关阅读:
    codeforces 484D D. Kindergarten(dp)
    codeforces 484B B. Maximum Value(二分)
    codeforces 484A A. Bits(贪心)
    51nod-1537 1537 分解(矩阵快速幂+找规律)
    大数取模
    小明的烦恼
    子网掩码
    How Many Tables
    N的N次方
    外星人的供给站
  • 原文地址:https://www.cnblogs.com/WizardCowboy/p/7643572.html
Copyright © 2020-2023  润新知