[codeforces] 877E. Danil and a Part-time Job(DFS序+线段树)

[codeforces] 877E. Danil and a Part-time Job(DFS序+线段树)

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题目链接: E. Danil and a Part-time Job
题目大意:
给一棵大小为n的以1为根节点的树。 每个点的值为0或1， 有q个操作:
1. pow v, 对于v以及v的子树所有节点01反转。
2. get v, 查询v以及v的子树中1节点的个数。

数据范围:
(1≤n,q≤2e5)

解题思路:
dfs序+线段树， 线段树用来维护每个点被反转了几次， 通过这道题让我又深一步的理解了线段树下放的思想： 当你查询到一个点的时候， 这个点肯定已经是被操作过的。所以反转应该写在下放的时候。

AC代码:

/******************************************** *Author*        :ZZZZone *Created Time*  : 五 11/ 3 20:03:46 2017 * Ended  Time*  : 六 11/ 4 00:05:58 2017*********************************************/#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>#include <stack>using namespace std;typedef pair<int, int> PII;typedef long long LL;typedef unsigned long long ULL;const int MaxN = 2e5;vector<int> edge[MaxN + 5];int head[MaxN + 5], tail[MaxN + 5], re[MaxN + 5];int n, q, tot, now;int a[MaxN + 5];int sum1[4 * MaxN + 5], sum0[4 * MaxN + 5], add[4 * MaxN + 5];void dfs(int u, int fa){    head[u] = ++tot;    re[tot] = u;    for(int i = 0; i < edge[u].size(); i++){        int v = edge[u][i];        if(v != fa) dfs(v, u);    }    tail[u] = tot;}void pushup(int rt){    sum1[rt] = sum1[rt << 1] + sum1[rt << 1 | 1];    sum0[rt] = sum0[rt << 1] + sum0[rt << 1 | 1];}void  build(int l, int r, int rt){    if(l == r){        now++;        if(a[re[now]] == 1) sum1[rt]++;        else sum0[rt]++;        return;    }    int mid = (l + r) >> 1;    build(l, mid, rt << 1);    build(mid + 1, r, rt << 1 | 1);    pushup(rt);}void pushdown(int rt){    if(add[rt]){        add[rt << 1] ^= 1;        add[rt << 1 | 1] ^= 1;        swap(sum1[rt << 1], sum0[rt << 1]);        swap(sum1[rt << 1 | 1], sum0[rt << 1 | 1]);        add[rt] = 0;    }}void update(int L, int R, int l, int r, int rt){    if(L <= l && R >= r){        swap(sum0[rt], sum1[rt]);        add[rt] ^= 1;        return;    }    pushdown(rt);        int mid = (l + r) >> 1;    if(L <= mid) update(L, R, l, mid, rt << 1);    if(R > mid) update(L, R,  mid + 1, r, rt << 1 | 1);    pushup(rt);}int query(int L, int R, int l, int r, int rt){    if(L <= l && R >= r){        //printf("%d %d %d   ", sum0[rt], sum1[rt], add[rt]);        return sum1[rt];    }    pushdown(rt);    int mid = (r + l) >> 1;    LL ret = 0;    if(L <= mid) ret += query(L, R, l, mid, rt << 1);    if(R > mid) ret += query(L, R, mid + 1, r, rt << 1 | 1);    return ret;}int main(){    scanf("%d", &n);    for(int i = 2; i <= n; i++){        int u;        scanf("%d", &u);        edge[u].push_back(i);    }    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);    dfs(1, 0);    scanf("%d", &q);    build(1, n, 1);    while(q--){        char tmp[5];        int p;        scanf("%s %d\n", tmp, &p);        if(tmp[0] == 'p') update(head[p], tail[p], 1, n, 1);        else if(tmp[0] == 'g') printf("%d\n", query(head[p], tail[p], 1, n, 1));    }    return 0;}