poj 3321 Apple Tree 线段树

一棵多叉树，每一个节点上都可以长出或摘掉一个苹果，节点从1至n标记，根为1节点

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn = 100010;
struct node{       //每个节点其子树在遍历后顺序的起始位置，即询问此节点及其子树时，线段树中对应的lef与rig位置
int lef, rig, id; // 遍历后坐标
}tree[maxn]; //yuan
int root[maxn]; // root[原] = 遍历后坐标
int n, q, v;
struct E{      //邻接表存树
int v, next;
}edge[maxn];
int pre[maxn], apple[maxn]; //apple记录次节点之前有误苹果
struct SEG{
int lef, rig, mid, ans;
}seg[maxn*4];
void init(){
int i, j, k, aa, bb, value = 0;
memset( pre, -1, sizeof(pre));
for( i= 1; i<n; i++){
scanf("%d%d", &aa, &bb);
edge[value].v = bb;
edge[value].next = pre[aa];
pre[aa] = value++;
}
for( i= 1; i<= n; i++){
apple[i] = 1;
}
}
void dfs(int pp){//后序遍历树
int i, j, k;
tree[pp].lef = v;
for( i= pre[pp]; i!= -1; i= edge[i].next){
dfs(edge[i].v);
}
root[pp] = v;
tree[pp].rig = v++;
}
void maketree(int num, int lef, int rig){
seg[num].lef = lef;
seg[num].rig = rig;
seg[num].mid = (lef + rig) >> 1;
seg[num].ans = rig - lef + 1;
if( lef != rig){
maketree(num*2, lef, seg[num].mid);
maketree(num*2+1, seg[num].mid+1, rig);
}
}
int cal( int num, int lef, int rig){
if( seg[num].lef == lef && seg[num].rig == rig){
return seg[num].ans;
}
if( rig <= seg[num].mid){
return cal( num*2, lef, rig);
}
else if( lef > seg[num].mid){
return cal( num*2+1 , lef, rig);
}
else return cal( num*2, lef , seg[num].mid) + cal(num*2+1, seg[num].mid+1, rig);
}
void insert( int num, int lef, int val){
seg[num].ans += val;
if( seg[num].lef == seg[num].rig) return;
if( lef <= seg[num].mid){
insert( num*2, lef, val);
}
else if( lef > seg[num].mid){
insert( num*2+1, lef, val);
}
}
int main(){
//freopen("1.txt", "r", stdin);
int i, j, k, aa, bb;
char tmp[10];
while( scanf("%d", &n) != EOF){
init();
v = 1;
dfs(1);
maketree(1, 1, n);
scanf("%d", &q);
for( i= 0; i<q; i++){
scanf("%s%d", tmp, &aa);
if( tmp[0] == 'Q'){
printf("%d\n", cal(1, tree[aa].lef, tree[aa].rig));
}
else{
if( apple[root[aa]] == 1) bb = -1; //注意，读入为原始编号，但线段树中为遍历后编号
else bb = 1;
insert( 1, root[aa], bb);
apple[root[aa]] = 1 - apple[root[aa]];
}
}
}
return 0;
}