POJ - 3321 Apple Tree

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1.题面

http://poj.org/problem?id=3321

2.题意

给你一个指定树根的树，每个节点上有一个为０或1的值，提供两种操作，1.是改变某个节点的值，在0,1之间进行转化，2.是询问某个节点子树中所有节点值之和。

3.思路

基本思想是一个子树的dfs时间戳在dfs序中是连接的。

这道题错了几发，因为没考虑到边的数量是点的两倍，没有开够节点。

4.代码

/*****************************************************************    > File Name: Cpp_Acm.cpp    > Author: Uncle_Sugar    > Mail: uncle_sugar@qq.com    > Created Time: 2016年07月18日 星期一 11时56分56秒*****************************************************************/# include <cstdio># include <cstring># include <cctype># include <cmath># include <cstdlib># include <climits># include <iostream># include <iomanip># include <set># include <map># include <vector># include <stack># include <queue># include <algorithm>using namespace std;const int debug = 1;const int size  = 100000 + 100; const int INF = INT_MAX>>1;typedef long long ll;int n,m,tmp;struct edge{int to,nxt;}e[2*size];int e_tot;int head[size];void AddEdge(int u,int v){e[e_tot].to = v;e[e_tot].nxt = head[u];head[u] = e_tot++;e[e_tot].to = u;e[e_tot].nxt = head[v];head[v] = e_tot++;}int dfsn[size],sz[size];int dfs_clock;void dfs(int fa,int n){dfsn[n] = ++dfs_clock;for (int i=head[n];i!=-1;i=e[i].nxt){if (e[i].to!=fa){dfs(n,e[i].to);sz[n] += sz[e[i].to];}} sz[n]++;}int BIT[size];inline int lowbit(int x){return (x - (x&(x-1)));}void Add(int p,int v){while (p<=n){BIT[p] += v;p += lowbit(p);}}int Sum(int p){int ret = 0;while (p>0){ret += BIT[p];p -= lowbit(p);}return ret;}int rd[size];int main(){int i,j;char cmd[10];while (scanf("%d",&n)!=EOF){if (n>100000)while (1);dfs_clock = 0;e_tot = 0;memset(sz,0,sizeof(sz));memset(head,-1,sizeof(head));memset(BIT,0,sizeof(BIT));for (i=0;i<n-1;i++){int a,b;scanf("%d%d",&a,&b);AddEdge(a,b);}dfs(-1,1);for (i=1;i<=n;i++){Add(dfsn[i],1);rd[i] = 1;}scanf("%d",&m);while (m--){scanf("%s%d",cmd,&tmp);if (cmd[0]=='C'){if (rd[tmp]==1){rd[tmp] -= 1;Add(dfsn[tmp],-1);}else {rd[tmp]+= 1;Add(dfsn[tmp],1);}}else {int k = Sum(dfsn[tmp]+sz[tmp]-1) - Sum(dfsn[tmp]-1);printf("%d\n",k);}}}return 0;}

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