CodeForces

题目链接：http://codeforces.com/problemset/problem/384/E

大意：给定一颗n节点的树，一共有两种操作：1. 给节点x加val，同时，它的所有孩子要减val，孙子加val，孙子的孩子减val，以此类推。2.查询节点x的值。

思路：对于某个节点来说，如果某节点的深度的奇偶性和该节点相同，则他们修改时符号也相同（要么都加要么都减），若奇偶性不同，则修改时符号也不同。因此，我们按照节点的奇偶性建立两棵线段树。修改时，分别修改在同一棵树上的节点和另一棵树上的节点即可。

#include<cstdio>#include<cstring>#include<string>#include<cctype>#include<iostream>#include<set>#include<map>#include<cmath>#include<sstream>#include<vector>#include<stack>#include<queue>#include<algorithm>#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define fin(a) freopen("a.txt","r",stdin)#define fout(a) freopen("a.txt","w",stdout)typedef long long LL;using namespace std;typedef pair<int, int> P;const int INF = 1e8 + 10;const int maxn = 2e5 + 10;int val[maxn];vector<int> G[maxn];int L[maxn], R[maxn];int cnt;bool vis[maxn];int depth[maxn];void dfs(int u, int d) {  L[u] = ++cnt; vis[u] = 1;  depth[u] = (d & 1);  for(int i = 0; i < G[u].size(); i++) {    int v = G[u][i];    if(vis[v]) continue;    dfs(v, d+1);  }  R[u] = cnt;}struct Tree {    int sum[maxn<<2], add[maxn<<2];    void PushDown(int rt, int m) {          if(add[rt]) {            add[rt<<1] += add[rt];            add[rt<<1|1] += add[rt];            sum[rt<<1] += add[rt] * (m - (m >> 1));            sum[rt<<1|1] += add[rt] * (m >> 1);            add[rt] = 0;          }    }    void build(int l, int r, int rt) {          if(l == r) {            sum[rt] = add[rt] = 0;          }          else {            int m = (l + r) >> 1;            build(lson);            build(rson);            sum[rt] = add[rt] = 0;          }    }    void update(int ql, int qr, int x, int l, int r, int rt) {          if(ql <= l && r <= qr) {            add[rt] += x;            sum[rt] += (r-l+1)*x;          }          else {            PushDown(rt, r-l+1);            int m = (l + r) >> 1;            if(ql <= m) update(ql, qr, x, lson);            if(m < qr) update(ql, qr, x, rson);            sum[rt] = sum[rt<<1] + sum[rt<<1|1];          }    }    int query(int p, int l, int r, int rt) {          if(l == r) return sum[rt];          PushDown(rt, r-l+1);          int m = (l + r) >> 1;          if(p <= m) return query(p, lson);          return query(p, rson);    }}tree[2];int main() {     int n, q;     scanf("%d%d", &n, &q);     for(int i = 1; i <= n; i++) {        scanf("%d", &val[i]);        G[i].clear();     }     for(int i = 1; i <= n-1; i++) {        int u, v;        scanf("%d%d", &u, &v);        G[u].push_back(v);        G[v].push_back(u);     }     memset(vis, false, sizeof vis);     dfs(1, 0);     tree[0].build(1, cnt, 1);     tree[1].build(1, cnt, 1);     for(int i = 1; i <= n; i++)        tree[depth[i]].update(L[i], L[i], val[i], 1, cnt, 1);     while(q--) {        int x, y, p;        scanf("%d", &p);        if(p == 1) {          scanf("%d%d", &x, &y);          int l = L[x], r = R[x];          tree[depth[x]].update(l, r, y, 1, cnt, 1);          tree[depth[x]^1].update(l, r, -y, 1, cnt, 1);        }        else {          scanf("%d", &x);          int p = L[x];          printf("%d\n", tree[depth[x]].query(p, 1, cnt, 1));        }     }  return 0;}