題干：

There is a company that has N employees(numbered from 1 to N),every employee in the company has a immediate boss (except for the leader of whole company).If you are the immediate boss of someone,that person is your subordinate, and all his subordinates are your subordinates as well. If you are nobody's boss, then you have no subordinates,the employee who has no immediate boss is the leader of whole company.So it means the N employees form a tree.?

The company usually assigns some tasks to some employees to finish.When a task is assigned to someone,He/She will assigned it to all his/her subordinates.In other words,the person and all his/her subordinates received a task in the same time. Furthermore,whenever a employee received a task,he/she will stop the current task(if he/she has) and start the new one.?

Write a program that will help in figuring out some employee’s current task after the company assign some tasks to some employee.

Input

The first line contains a single positive integer T( T <= 10 ), indicates the number of test cases.?

For each test case:?

The first line contains an integer N (N ≤ 50,000) , which is the number of the employees.?

The following N - 1 lines each contain two integers u and v, which means the employee v is the immediate boss of employee u(1<=u,v<=N).?

The next line contains an integer M (M ≤ 50,000).?

The following M lines each contain a message which is either?

"C x" which means an inquiry for the current task of employee x?

or?

"T x y"which means the company assign task y to employee x.?

(1<=x<=N,0<=y<=10^9)

Output

For each test case, print the test case number (beginning with 1) in the first line and then for every inquiry, output the correspond answer per line.

Sample Input

1 5 4 3 3 2 1 3 5 2 5 C 3 T 2 1C 3 T 3 2 C 3

Sample Output

Case #1: -1 1 2

題目大意：

先輸入n個人，然后是n個人的關系(下屬 上司)，然后分配任務，每一個上司分到一個任務就把任務給他的所有下屬做，即他的下屬的任務都是這個任務，查詢一個人該時刻做的任務。

解題報告：

可以建模一下：是給定一棵樹，然后一種操作是指定一個點，這個點及這個點的的子樹被染色，另一種操作是指定一個點，問這個點的顏色。dfs序的實質就是：通過dfs樹將這棵樹放在線段上，然后用線段樹優(yōu)化求解。因為是單點查詢所以不需要pushup，因為是區(qū)間更新并且需要葉子結點的值，所以需要pushdown。

AC代碼：

#include<bits/stdc++.h>using namespace std; const int MAX = 50000 + 5; int n; int cnt,id; int head[MAX*2]; int q[MAX*4],in[MAX*4],out[MAX*4],rudu[MAX*4]; struct Edge {int to;int ne; } e[MAX*4];//要乘2 嗎？ struct TREE {int l,r;int val;int laz; } tree[MAX * 10];void dfs(int x,int root) {q[++id] = x;in[x] = id;for(int i = head[x]; i!=-1; i = e[i].ne) {dfs(e[i].to,x);}out[x] = id; } void add(int u,int v) {e[++cnt].to = v;e[cnt].ne = head[u];head[u] = cnt; } void build(int l,int r,int cur) {tree[cur].l = l;tree[cur].r = r;tree[cur].val = -1;tree[cur].laz = 0;//懶標記初始化成0 if(l == r) {return ;}int m = (l + r) /2;build(l,m,cur*2);build(m+1,r,cur*2+1); } void pushup(int cur) {if(tree[cur*2].val == tree[cur*2+1].val) tree[cur].val = tree[cur*2].val;else tree[cur].val = -2;//染色問題 通解？ 但是此題是單點查詢所以不需要pushup？ } void pushdown(int cur) {if(tree[cur].laz == 0) return ;tree[cur*2].val =tree[cur*2].laz = tree[cur*2+1].laz = tree[cur*2+1].val = tree[cur].laz;tree[cur].laz = 0; } void update(int pl,int pr, int val,int cur) {if(pl <= tree[cur].l && pr >= tree[cur].r) {tree[cur].val = val;tree[cur].laz = val;return ;}pushdown(cur);if(pl <= tree[cur*2].r) update(pl,pr,val,2*cur);if(pr >= tree[cur*2+1].l) update(pl,pr,val,2*cur+1);//pushup(cur); } int query(int tar,int cur) {if(tree[cur].l == tree[cur].r) {return tree[cur].val;}pushdown(cur);if(tar <= tree[cur*2].r) return query(tar,cur*2);else return query(tar,cur*2+1);//pushup(cur); } int main() { // freopen("in.txt","r",stdin);int t;int m,iCase = 0;int u,v;char op[10];cin>>t;while(t--) {printf("Case #%d:\n",++iCase);cnt = id = 0;memset(head,-1,sizeof(head));memset(rudu,0,sizeof rudu);memset(e,0,sizeof(e));scanf("%d",&n);for(int i = 1; i<n; i++) {//讀入n-1次 scanf("%d%d",&u,&v);add(v,u);rudu[u]++;} int st =1;for(int i = 1; i<=n; i++) {if(rudu[i] == 0) {st = i;break;}} dfs(st,0);build(1,n,1);scanf("%d",&m);while(m--) {scanf("%s",op);if(op[0] == 'C') {scanf("%d",&u);printf("%d\n",query(in[u],1));} else {scanf("%d%d",&u,&v);//把以u為根節(jié)點的子樹更改為v update(in[u],out[u],v,1);}}}return 0 ; } //19:15

總結：

? ?注意dfs序，必須從根節(jié)點開始，所以還需要找根節(jié)點！！！這里是用拓撲排序中入度的思想找的根節(jié)點。?

??【POJ - 3321】 Apple Tree這道題也是dfs序，但是沒有找根節(jié)點，因為題目告訴你了1號節(jié)點是根節(jié)點.

本題的數組都是佛系開的，其實只是tree需要4倍，其他的都開MAX就夠用。RE了一次是因為，發(fā)現(xiàn)dfs需要找根節(jié)點而不能直接用1號節(jié)點之后，加了rudu數組，但是rudu沒有每次初始化，所以re了。。（話說不應該wa嗎，搞不懂。。

?

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