題意：

染色問題：給一個固定結構的樹，現在有兩個操作：
（1） y 將結點x及其所有后代結點染成顏色y；
（2）查詢結點x當前的顏色。
其實就是區間染色問題，不過需要dfs預處理，

題目：

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 1
C 3
T 3 2
C 3

Sample Output

Case #1:
-1
1
2

分析：

1.由于題目給的是一個固定的樹，所以我們需要對它進行處理，使之成為我們好處理的區間問題。
2.關于預處理我采用的是先用vector容器進行存邊，之后dfs，用兩個數組分別存儲以某點為“”祖宗節點“”的區間起始（類似時間戳的思想，將某個點的編號）
3.預處理之后就是線段樹模板啦嘿嘿，這個沒啥好說的，這里不再贅述： 區間修改+單點查詢

AC代碼：

#include<stdio.h> #include<string.h> #include<vector> #include<algorithm> using namespace std; const int M=5e4+10; int t,n,m,ans,a,b,k; vector<int>ve[M]; int L[M],R[M],dp[M<<2],lazy[M<<2]; bool vis[M]; char s[5]; void dfs(int x) {L[x]=++ans;for(int i=0; i<ve[x].size(); i++)dfs(ve[x][i]);R[x]=ans; } void pushdown(int x) {if(lazy[x]){lazy[x<<1]=lazy[x<<1|1]=lazy[x];dp[x<<1]=dp[x<<1|1]=dp[x];lazy[x]=0;} } void update(int mi,int ma,int l,int r,int x,int value) {if(l>=mi&&r<=ma){dp[x]=lazy[x]=value;return;}pushdown(x);int mid=(l+r)>>1;if(mid>=mi)update(mi,ma,l,mid,x<<1,value);if(mid<ma)update(mi,ma,mid+1,r,x<<1|1,value); } int query(int now,int l,int r,int x)//單點修改 {if(l==r)return dp[x];pushdown(x);int mid=(l+r)>>1;if(mid>=now)return query(now,l,mid,x<<1);elsereturn query(now,mid+1,r,x<<1|1); } int main() {scanf("%d",&t);k=0;while(t--){ans=0;memset(dp,-1,sizeof(dp));memset(lazy,0,sizeof(lazy));memset(vis,false,sizeof(vis));scanf("%d",&n);for(int i=1; i<=n; i++)ve[i].clear();for(int i=0; i<n-1; i++){int u,v;scanf("%d%d",&u,&v);ve[v].push_back(u);//接下來的N-1行每行包含兩個整數u和v，這表示雇員v是雇員u（1 <= u，v <= N）的直接老板。vis[u]=true;}for(int i=1; i<=n; i++)if(!vis[i])dfs(i);scanf("%d",&m);printf("Case #%d:\n",++k);while(m--){scanf("%s",s);if(s[0]=='C'){scanf("%d",&a);printf("%d\n",query(L[a],1,ans,1));}else{scanf("%d%d",&a,&b);update(L[a],R[a],1,ans,1,b);}}}return 0; }

備戰ing，題目分析簡略，見諒，轉載請注明出處。。。。。

總結

以上是生活随笔為你收集整理的Assign the task HDU - 3974（线段树+dfs建树+单点查询+区间修改）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。