題干：

There is an apple tree outside of kaka's house. Every autumn, a lot of apples will grow in the tree. Kaka likes apple very much, so he has been carefully nurturing the big apple tree.

The tree has?N?forks which are connected by branches. Kaka numbers the forks by 1 to?N?and the root is always numbered by 1. Apples will grow on the forks and two apple won't grow on the same fork. kaka wants to know how many apples are there in a sub-tree, for his study of the produce ability of the apple tree.

The trouble is that a new apple may grow on an empty fork some time and kaka may pick an apple from the tree for his dessert. Can you help kaka?

?

?

Input

The first line contains an integer?N?(N?≤ 100,000) , which is the number of the forks in the tree.
The following?N?- 1 lines each contain two integers?u?and?v, which means fork?u?and fork?v?are connected by a branch.
The next line contains an integer?M?(M?≤ 100,000).
The following?M?lines each contain a message which is either
"C?x" which means the existence of the apple on fork?x?has been changed. i.e. if there is an apple on the fork, then Kaka pick it; otherwise a new apple has grown on the empty fork.
or
"Q?x" which means an inquiry for the number of apples in the sub-tree above the fork?x, including the apple (if exists) on the fork x
Note the tree is full of apples at the beginning

Output

For every inquiry, output the correspond answer per line.

Sample Input

3 1 2 1 3 3 Q 1 C 2 Q 1

Sample Output

3 2

題目大意：

給出一個蘋果樹，每個節點一開始都有蘋果

C X，如果X點有蘋果，則拿掉，如果沒有，則新長出一個（即：可以用異或處理）

Q X，查詢X點與它的所有后代分支一共有幾個蘋果

解題報告：

? ? ?dfs序，然后用線段樹單點更新+區間查詢一下就可以了。（dfs序是專門用來處理一棵樹的子樹查詢的，需要用線段樹或樹狀數組維護查詢）

AC代碼：（用輸入外掛，用時少了200ms左右，不知道rank1的63ms是怎么做到的）

#include<cstdio> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> using namespace std; const int MAX = 100000 + 5; int n,m; int cnt,id; int head[MAX*10]; int q[MAX*4],in[MAX*4],out[MAX*4]; struct Edge {int fr;int to;int ne; } e[MAX * 4];//因為可能雙向邊？ struct TREE {int l,r;int val; } tree[MAX * 8]; void add(int u,int v) {e[++cnt].to = v;e[cnt].fr = u;e[cnt].ne = head[u];head[u] = cnt; } void pushup(int cur) {tree[cur].val = tree[cur*2].val + tree[cur*2+1].val; } void build(int l,int r,int cur) {tree[cur].l = l;tree[cur].r = r;if(l == r) {tree[cur].val = 1;return ;//又忘寫return了。。。 }int m = (l+r)/2;//剛開始沒寫build遞歸。。。 build(l,m,cur*2);build(m+1,r,cur*2+1);pushup(cur); } void dfs(int cur,int root) //cur為當前點，root為根節點 {q[++id] =cur;in[cur] = id;for(int i = head[cur]; i!=-1;i = e[i].ne) {dfs(e[i].to,cur);}out[cur] = id; } void update(int tar,int cur) {if(tree[cur].l == tree[cur].r) {tree[cur].val ^= 1;return ; }if(tar <= tree[cur*2].r) update(tar,cur*2);//剛開始寫成tree[cur*2].l了else update(tar,cur*2+1);pushup(cur); } int query(int pl,int pr,int cur) {if(pl <= tree[cur].l && pr >= tree[cur].r) {return tree[cur].val;}int res = 0;if(pl <= tree[cur*2].r) res += query(pl,pr,cur*2);if(pr >= tree[cur*2+1].l) res += query(pl,pr,cur*2+1);return res; } int read() {int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f; } int main() {cin>>n;char op[10];int x;cnt = id = 0;memset(head,-1,sizeof head);int u,v;for(int i = 1; i<n; i++) {//scanf("%d%d",&u,&v);u = read();v = read();add(u,v);}dfs(1,0);//這個0 是沒用的 build(1,n,1); // printf("**********\n"); // for(int i = 1; i<=5; i++) { // printf("%d| ",tree[i].val); // } // printf("\n****************\n");scanf("%d",&m);while(m--) {scanf("%s",op);if(op[0] == 'C') {//scanf("%d",&x);x = read();update(in[x],1);}else {//scanf("%d",&x);x = read();int ans = query(in[x],out[x],1);printf("%d\n",ans);}} return 0 ;}

總結：

? ?小錯誤還是很多啊。。。

總結

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