TmonoのACM
坚持是一种饼~

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:

  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 }
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posted on 2014-09-12 18:36  Pobo_biu  阅读(282)  评论(0编辑  收藏  举报