当前位置 博文首页 > danzh:洛谷P2146 [NOI2015]软件包管理器 树剖+线段树+区间覆盖

    danzh:洛谷P2146 [NOI2015]软件包管理器 树剖+线段树+区间覆盖

    作者:[db:作者] 时间:2021-09-21 20:59


    洛谷P2146 [NOI2015]软件包管理器


    标签

    • 树链剖分
    • 线段树
    • 区间覆盖

    简明题意

    • 给一棵树,支持两种操作
      1. 询问树上1-u路径中点权为0的个数,并将他们改为1
      2. 询问u的子树中点权为1的个数,并将它们改为0

    思路

    • 直接上树剖啊,太简单了。这里简单说一下线段树吧,既然是个区间操作,肯定来一个带tag的线段树,然后我们发现,这个tag是覆盖型的,所以下传标记时直接覆盖原标记就可以了,非常简单

    注意事项


    总结


    AC代码

    #include<cstdio>
    #include<vector>
    #include<algorithm>
    using namespace std;
    
    const int maxn = 1e5 + 10;
    
    int n, m;
    vector<int> g[maxn];
    
    int dep[maxn], son[maxn], siz[maxn], fa[maxn];
    void dfs1(int u, int f, int deep)
    {
       fa[u] = f;
       dep[u] = deep;
       siz[u] = 1;
    
       int max_son = -1;
       for (auto& v : g[u])
       	if (v != f)
       	{
       		dfs1(v, u, deep + 1);
       		siz[u] += siz[v];
       		if (siz[v] > max_son)
       			max_son = siz[v], son[u] = v;
       	}
    }
    
    int id[maxn], cnt, top[maxn], w[maxn];
    void dfs2(int u, int topf)
    {
       id[u] = ++cnt;
       top[u] = topf;
    
       if (son[u])
       {
       	dfs2(son[u], topf);
       	for (auto& v : g[u])
       		if (v != fa[u] && v != son[u])
       			dfs2(v, v);
       }
    }
    
    struct Node
    {
       int l, r, sum;
       int tag;
    };
    
    Node tree[maxn * 4];
    
    void build(int o, int l, int r)
    {
       tree[o].l = l, tree[o].r = r;
       tree[o].tag = -1;
       if (l == r)
       	return;
    
       int mid = (l + r) / 2;
       build(o * 2, l, mid);
       build(o * 2 + 1, mid + 1, r);
    }
    
    void spread(int o)
    {
       if (tree[o].tag != -1)
       {
       	tree[o].sum = tree[o].tag == 1 ? (tree[o].r - tree[o].l + 1) : 0;
       	if (tree[o].l != tree[o].r)
       		tree[o * 2].tag = tree[o * 2 + 1].tag = tree[o].tag;
       	tree[o].tag = -1;
       }
    }
    
    void update(int o)
    {
       if (tree[o].l != tree[o].r)
       {
       	spread(o * 2), spread(o * 2 + 1);
       	tree[o].sum = tree[o * 2].sum + tree[o * 2 + 1].sum;
       }
    }
    
    void change(int o, int l, int r, int c)
    {
       spread(o);
    
       if (tree[o].l == l && tree[o].r == r)
       {
       	tree[o].tag = c;
       	spread(o);
       	return;
       }
    
       int mid = (tree[o].l + tree[o].r) / 2;
       if (r <= mid)
       	change(o * 2, l, r, c);
       else if (l > mid)
       	change(o * 2 + 1, l, r, c);
       else
       	change(o * 2, l, mid, c), change(o * 2 + 1, mid + 1, r, c);
    
       update(o);
    }
    
    int ask(int o, int l, int r, int stau)
    {
       spread(o);
    
       if (tree[o].l == l && tree[o].r == r)
       	return stau == 1 ? tree[o].sum : tree[o].r - tree[o].l + 1 - tree[o].sum;
    
       int mid = (tree[o].l + tree[o].r) / 2;
       if (r <= mid)
       	return ask(o * 2, l, r, stau);
       else if (l > mid)
       	return ask(o * 2 + 1, l, r, stau);
       else
       	return ask(o * 2, l, mid, stau) + ask(o * 2 + 1, mid + 1, r, stau);
    }
    
    void solve()
    {
       scanf("%d", &n);
       for (int i = 2; i <= n; i++)
       {
       	int f;
       	scanf("%d", &f);
       	f++;
    
       	g[f].push_back(i), g[i].push_back(f);
       }
    
       dfs1(1, 1, 1);
       dfs2(1, 1);
    
       build(1, 1, n);
    
       scanf("%d", &m);
       while (m--)
       {
       	char cmd[20];
       	int u;
       	scanf("%s%d", cmd, &u);
       	u++