题意:一棵根为1的多叉树有n个点,题目有m次询问。第一行输入n和m,第二行输入n-1条边, 以后m行输入操作,操作有两种:1 x val 表示 节点的值x+val,同时它的儿子层节点的值-val,孙子层节点的值+val...如此往下直到叶子节点;2 x 表示输出x节点的当前值。
思路:类似poj3321,用dfs序重新表示每个节点,这样更新子树的操作就变成更新区间了,区间是:[i, i+cnt]【当前节点的dfs序为 i, 儿子数为cnt】,查询同理,单点查询当前节点的dfs序。但是这道题的dfs序,奇、偶层的节点要分开来记录,也要建奇、偶两棵线段树。dfs过程中,还有每次更新查询都要判断在哪层。具体细节看代码。
AC代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstdio> 3 #include <vector> 4 #include <algorithm> 5 #include <cstring> 6 using namespace std; 7 #define maxn 201000 8 #define lson l, m, rt<<1 9 #define rson m+1, r, rt<<1|1 10 #define ll long long 11 struct node 12 { 13 int cnt_o, cnt_e; //分别记录奇层、偶层有几个儿子 14 int val; 15 int num, dep, next; // num记录dfs序,dep 记录节点深度(这里节点1深度为1),next记录下一层的开头的节点。 16 }arr[maxn]; 17 int sgt_o[maxn<<2], sgt_e[maxn<<2]; 18 int lazy_o[maxn<<2], lazy_e[maxn<<2]; 19 vector<int> odd, even; //分别记录在奇、偶层的原节点序号 20 int vis[maxn]; 21 int n, m; 22 int next[maxn<<1], first[maxn], v[maxn<<1], u[maxn<<1]; //邻接表 23 void init() 24 { 25 odd.clear(); even.clear(); 26 memset(first, -1, sizeof(first)); 27 memset(next, 0, sizeof(next)); 28 memset(vis, 0, sizeof(vis)); 29 for(int i = 1; i <= n; i++) { 30 scanf("%d", &arr[i].val); 31 arr[i].cnt_e = arr[i].cnt_o = arr[i].next = 0; 32 } 33 for(int i = 0; i < n-1; i++) { //用邻接表存图 34 scanf("%d%d", &u[i], &v[i]); 35 next[i] = first[u[i]]; 36 first[u[i]] = i; 37 u[n+i] = v[i]; 38 v[n+i] = u[i]; 39 next[n+i] = first[v[i]]; 40 first[v[i]] = n+i; 41 } 42 } 43 struct child //记录奇、偶的儿子数 44 { 45 int e, o; 46 }; 47 child dfs(int i, int &num1, int &num2, int dep) //num1是奇层的dfs序,num2是偶层的dfs序 48 { 49 if(dep&1) odd.push_back(i); 50 else even.push_back(i); 51 52 if(dep&1) arr[i].num = num1++; 53 else arr[i].num = num2++; 54 arr[i].dep = dep; 55 56 vis[i] = 1; 57 58 int flag = 0; 59 if(dep&1) 60 for(int e = first[i]; e != -1; e = next[e]) { 61 if(!vis[v[e]]) { 62 if(!flag) { 63 arr[i].next = v[e]; flag = 1; 64 } 65 child x = dfs(v[e], num1, num2, dep+1); 66 arr[i].cnt_e += x.e; 67 arr[i].cnt_o += x.o; 68 } 69 } 70 else { 71 for(int e = first[i]; e != -1; e = next[e]) { 72 if(!vis[v[e]]) { 73 if(!flag) { 74 arr[i].next = v[e]; flag = 1; 75 } 76 child x = dfs(v[e], num1, num2, dep+1); 77 arr[i].cnt_e += x.e; 78 arr[i].cnt_o += x.o; 79 } 80 } 81 } 82 child xx; xx.o = arr[i].cnt_o; xx.e = arr[i].cnt_e; 83 if(dep&1) xx.o++; else xx.e++; 84 return xx; 85 } 86 87 void build_o(int l, int r, int rt) //建奇层节点的线段树 88 { 89 sgt_o[rt] = 0; 90 if(l == r) { 91 int x = odd[l-1]; 92 sgt_o[rt] = arr[x].val; 93 return; 94 } 95 int m = (r+l)>>1; 96 build_o(lson); 97 build_o(rson); 98 } 99 void build_e(int l, int r, int rt) //建偶层节点的线段树 100 { 101 sgt_e[rt] = 0; 102 if(l == r) { 103 int x = even[l-1]; 104 sgt_e[rt] = arr[x].val; 105 return; 106 } 107 int m = (l+r)>>1; 108 build_e(lson); 109 build_e(rson); 110 } 111 void push_down(int rt, int x, int *lazy) 112 { 113 if(lazy[rt] != 0) { 114 lazy[rt<<1] += lazy[rt]; 115 lazy[rt<<1|1] += lazy[rt]; 116 lazy[rt] = 0; 117 } 118 } 119 void change(int l, int r, int rt, int L, int R, int del, int *sgt, int *lazy) 120 { 121 if(L <= l && r <= R) 122 { 123 lazy[rt] += del; 124 return; 125 } 126 int x = (r-l+1); 127 push_down(rt, x, lazy); 128 int m = (l+r)>>1; 129 if(L <= m) change(lson, L, R, del, sgt, lazy); 130 if(m < R) change(rson, L, R, del, sgt, lazy); 131 } 132 int query(int l, int r, int rt, int pos, int *sgt, int *lazy) 133 { 134 if(l == r) { 135 sgt[rt] += lazy[rt]; 136 lazy[rt] = 0; 137 return sgt[rt]; 138 } 139 int m = (l+r)>>1; 140 push_down(rt, r-l+1, lazy); 141 if(pos <= m) return query(lson, pos, sgt, lazy); 142 return query(rson, pos, sgt, lazy); 143 } 144 void work() 145 { 146 init(); 147 int num1 = 1, num2 = 1; 148 dfs(1, num1, num2, 1); 149 int n1 = odd.size(), n2 = even.size(); 150 if(n1 > 0)build_o(1, n1, 1); 151 if(n2 > 0)build_e(1, n2, 1); 152 while(m--) { 153 int a, b; scanf("%d%d", &a, &b); 154 if(a == 1) { 155 int c; scanf("%d", &c); 156 if(arr[b].dep&1) { 157 change(1, n1, 1, arr[b].num, arr[b].num+arr[b].cnt_o, c, sgt_o, lazy_o); 158 if(arr[b].next != 0) { 159 int ne = arr[b].next; 160 change(1, n2, 1, arr[ne].num, arr[ne].num+arr[b].cnt_e-1, -c, sgt_e, lazy_e); 161 } 162 } 163 else { 164 change(1, n2, 1, arr[b].num, arr[b].num+arr[b].cnt_e, c, sgt_e, lazy_e); 165 if(arr[b].next != 0) { 166 int ne = arr[b].next; 167 change(1, n1, 1, arr[ne].num, arr[ne].num+arr[b].cnt_o-1, -c, sgt_o, lazy_o); 168 } 169 } 170 } 171 else { 172 int res; 173 if(arr[b].dep&1) { 174 res = query(1, n1, 1, arr[b].num, sgt_o, lazy_o); 175 } 176 else { 177 res = query(1, n2, 1, arr[b].num, sgt_e, lazy_e); 178 } 179 printf("%d\n", res); 180 } 181 } 182 } 183 int main() 184 { 185 while(scanf("%d%d", &n, &m) != EOF) work(); 186 return 0; 187 }