zoukankan      html  css  js  c++  java
  • [HDU3487]Play with Chain

    Problem

    有n个数,为1~n。有两种操作:
    Cut x y z: 把x到y的区间切割下来后,放到改变后的序列的z位后
    Flip x y: 把x到y的区间翻转

    Solution

    splay模板题

    Notice

    注意0

    Code

    #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 = 300000;
    const double eps = 1e-6, phi = acos(-1);
    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, root, cnt, X[N + 5], Y[N + 5], n;
    struct node
    {
    	int val[N + 5], son[2][N + 5], parent[N + 5], Size[N + 5], rev[N + 5];
    	inline void up(int u)
    	{
    	    Size[u] = Size[son[0][u]] + Size[son[1][u]] + 1;
    	}
    	inline void down(int u)
    	{
    	    if (rev[u])
            {
                swap(son[0][u], son[1][u]);
                rev[son[0][u]] ^= 1, rev[son[1][u]] ^= 1;
                rev[u] = 0;
            }
    	}
    	void Newnode(int &u, int from, int v)
    	{
    	    u = ++point;
            val[u] = v;
            son[0][u] = son[1][u] = rev[u] = 0;
            parent[u] = from, Size[u] = 1;
    	}
        void Build(int &u, int l, int r, int from)
        {
            int mid = (l + r) >> 1;
            Newnode(u, from, X[mid]);
            if (l < mid) Build(son[0][u], l, mid - 1, u);
            if (r > mid) Build(son[1][u], mid + 1, r, u);
            up(u);
        }
        void Init(int n)
        {
            rep(i, 1, n) X[i] = i;
            point = 0, cnt = 0;
            son[0][0] = son[1][0] = Size[0] = val[0] = parent[0] = rev[0] = 0;
            Newnode(root, 0, -1);
            Newnode(son[1][root], root, -1);
            Build(son[0][son[1][root]], 1, n, son[1][root]);
            up(son[1][root]), up(root);
        }
    
    	void Rotate(int x, int &rt)
    	{
    		int y = parent[x], z = parent[y];
    		down(y), down(x);
    		int l = (son[1][y] == x), r = 1 - l;
    		if (y == rt) rt = x;
    		else if (son[0][z] == y) son[0][z] = x;
    		else son[1][z] = x;
    		parent[x] = z;
    		parent[son[r][x]] = y, son[l][y] = son[r][x];
    		parent[y] = x, son[r][x] = y;
    		up(y), up(x);
    	}
    	void Splay(int x, int &rt)
    	{
    		while (x != rt)
    		{
    			int y = parent[x], z = parent[y];
    			down(z), down(y);
    			if (y != rt)
    			{
    				if ((son[0][z] == y) ^ (son[0][y] == x))
    					Rotate(x, rt);
    				else Rotate(y, rt);
    			}
    			Rotate(x, rt);
    		}
    	}
    
        void Out(int u)
        {
            if (!u || val[u] == 0) return;
            down(u);
            Out(son[0][u]);
            if (val[u] != -1) Y[++cnt] = val[u];
            Out(son[1][u]);
        }
    	int Find(int u, int x)
    	{
    	    down(u);
    		if (x <= Size[son[0][u]]) return Find(son[0][u], x);
    		if (x > Size[son[0][u]] + 1) return Find(son[1][u], x - Size[son[0][u]] - 1);
    		return u;
    	}
    	void Split(int l, int r)
    	{
    	    int x = Find(root, l - 1 + 1);
    	    int y = Find(root, r + 1 + 1);
    	    Splay(x, root);
    	    Splay(y, son[1][root]);
    	}
    
        void Cut(int x, int y, int z)
        {
            Split(x, y);
            int tt = son[0][son[1][root]];
            son[0][son[1][root]] = 0;
            up(son[1][root]), up(root);
            Split(z + 1, z);
            son[0][son[1][root]] = tt;
            parent[tt] = son[1][root];
            up(son[1][root]), up(root);
        }
        void Flip(int x, int y)
        {
            Split(x, y);
            rev[son[0][son[1][root]]] ^= 1;
        }
    }Splay_tree;
    
    int sqz()
    {
        while(1)
        {
            n = read(); int q = read();
            if (n < 0 && q < 0) break;
            Splay_tree.Init(n);
            char op[10];
            int x, y, z;
            while (q--)
            {
                scanf("%s", op);
                if (op[0] == 'C')
                {
                    x = read(), y = read(), z = read();
                    Splay_tree.Cut(x, y, z);
                }
                else
                {
                    x = read(), y = read();
                    Splay_tree.Flip(x, y);
                }
            }
            Splay_tree.Out(root);
            rep(i, 1, n - 1) printf("%d ", Y[i]);
            printf("%d
    ", Y[n]);
        }
        return 0;
    }
    
  • 相关阅读:
    java 利用jsoup 爬取知乎首页问题
    ROIAlign, ROIPooling及ROIWarp对比
    yii2.0 gii
    mysql索引操作
    lbs basic mongodb
    php操作mongodb
    设计模式六大原则
    Java集合
    Java 快速失败( fail-fast ) 安全失败( fail-safe )
    计数数组中值的出现次数
  • 原文地址:https://www.cnblogs.com/WizardCowboy/p/7631420.html
Copyright © 2011-2022 走看看