• Codeforces 877E Danil and a Part-time Job(dfs序 + 线段树)


    题目链接   Danil and a Part-time Job

    题意    给出一系列询问或者修改操作

       $pow$ $x$表示把以$x$为根的子树的所有结点的状态取反($0$变$1$,$1$变$0$)

       $get$  $x$表示求以$x$为根的子树中状态为$1$的结点数。

     

    首先大力$dfs$序,然后线段树操作一下。

    具体问题转化为:区间翻转,区间求和。

    #include <bits/stdc++.h>
    
    using namespace std;
    
    #define rep(i, a, b)	for (int i(a); i <= (b); ++i)
    #define dec(i, a, b)	for (int i(a); i >= (b); --i)
    #define ls		i << 1
    #define	rs		i << 1 | 1
    #define lson		ls, L, mid
    #define	rson		rs, mid + 1, R
    
    const int N = 2e5 + 10;
    
    struct node{
    	int s[2];
    	int len;
    } t[N * 3];
    
    int in[N], out[N], dc[N * 3], dt[N * 3], f[N], a[N];
    int n, q, ti;
    
    vector <int> v[N];
    
    void dfs(int x){
    	in[x] = ++ti;
    	f[ti] = x;
    	for (auto u : v[x]) dfs(u);
    	out[x] = ti;
    }
    
    node update(node x, node y){
    	node ans;
    	ans.len = x.len + y.len;
    	rep(i, 0, 1) ans.s[i] = x.s[i] + y.s[i];
    	return ans;
    }
    
    void build(int i, int L, int R){
    	if (L == R){
    		int z = a[f[L]];
    		t[i].len = t[i].s[z] = 1;
    		t[i].s[z ^ 1] = 0;
    		dc[i] = -1;
    		dt[i] = 0;
    		return ;
    	}
    
    	int mid = (L + R) >> 1;
    	build(lson);
    	build(rson);
    	t[i] = update(t[ls], t[rs]);
    	dc[i] = -1;
    	dt[i] = 0;
    }
    
    void cover(int i, int z){
    	dc[i] = z;
    	dt[i] = 0;
    	t[i].s[z] = t[i].len;
    	t[i].s[z ^ 1] = 0;
    }
    
    void turn(int i, int z){
    	if (~z){
    		dc[i] ^= 1;
    		t[i].s[z] = 0;
    		t[i].s[z ^ 1] = t[i].len;
    	}
    
    	else{
    		dt[i] ^= 1;
    		swap(t[i].s[0], t[i].s[1]);
    	}
    }
    
    void pushdown(int i){
    	if (~dc[i]){
    		cover(ls, dc[i]);
    		cover(rs, dc[i]);
    		dc[i] = -1;
    	}
    
    	if (dt[i]){
    		turn(ls, dc[ls]);
    		turn(rs, dc[rs]);
    		dt[i] = 0;
    	}
    }
    
    void Turn(int i, int L, int R, int x, int y){
    	if (x <= L && R <= y){
    		turn(i, dc[i]);
    		return;
    	}
    
    	pushdown(i);
    	int mid = (L + R) >> 1;
    	if (x <= mid) Turn(lson, x, y);
    	if (y > mid)  Turn(rson, x, y);
    	t[i] = update(t[ls], t[rs]);
    }
    
    int query(int i, int L, int R, int x, int y){
    	int ret = 0;
    	if (x <= L && R <= y) return t[i].s[1];
    	int mid = (L + R) >> 1;
    	pushdown(i);
    	if (x <= mid) ret += query(lson, x, y);
    	if (y > mid ) ret += query(rson, x, y);
    	return ret;
    }
    
    int main(){
    
    	scanf("%d", &n);
    	rep(i, 2, n){
    		int x;
    		scanf("%d", &x);
    		v[x].push_back(i);
    	}
    
    	dfs(1);
    	rep(i, 1, n) scanf("%d", a + i);
    	build(1, 1, n);
    
    	for (scanf("%d", &q); q--; ){
    		char op[10]; int x;
    		scanf("%s%d", op, &x);
    		if (op[0] == 'g') printf("%d
    ", query(1, 1, n, in[x], out[x]));
    		else Turn(1, 1, n, in[x], out[x]);
    	}
    
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/cxhscst2/p/7762425.html
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