HDU 3974 Assign the task（简单线段树）

Assign the task

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 636    Accepted Submission(s): 322


Problem Description
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
 

 

Source
2011 Multi-University Training Contest 14 - Host by FZU
 

 

 

用线段树修改区间值，查询单点值。

好久没写线段树了，这都写挫。。。

  1 /* ***********************************************
  2 Author        :kuangbin
  3 Created Time  :2013-11-17 19:50:24
  4 File Name     :E:\2013ACM\比赛练习\2013-11-17\C.cpp
  5 ************************************************ */
  6 
  7 #include <stdio.h>
  8 #include <string.h>
  9 #include <iostream>
 10 #include <algorithm>
 11 #include <vector>
 12 #include <queue>
 13 #include <set>
 14 #include <map>
 15 #include <string>
 16 #include <math.h>
 17 #include <stdlib.h>
 18 #include <time.h>
 19 using namespace std;
 20 
 21 const int MAXN = 50010;
 22 struct Edge
 23 {
 24     int to,next;
 25 }edge[MAXN];
 26 int head[MAXN],tot;
 27 int cnt;
 28 int start[MAXN],end[MAXN];
 29 void init()
 30 {
 31     cnt = 0;
 32     tot = 0;
 33     memset(head,-1,sizeof(head));
 34 }
 35 void addedge(int u,int v)
 36 {
 37     edge[tot].to = v;
 38     edge[tot].next = head[u];
 39     head[u] = tot++;
 40 }
 41 void dfs(int u)
 42 {
 43     ++cnt;
 44     start[u] = cnt;
 45     for(int i = head[u];i != -1;i = edge[i].next)
 46     {
 47         dfs(edge[i].to);
 48     }
 49     end[u] = cnt;
 50 }
 51 struct Node
 52 {
 53     int l,r;
 54     int val;
 55     int lazy;
 56 }segTree[MAXN*4];
 57 void Update_Same(int r,int v)
 58 {
 59     if(r)
 60     {
 61         segTree[r].val = v;
 62         segTree[r].lazy = 1;
 63     }
 64 }
 65 void push_down(int r)
 66 {
 67     if(segTree[r].lazy)
 68     {
 69         Update_Same(r<<1,segTree[r].val);
 70         Update_Same((r<<1)|1,segTree[r].val);
 71         segTree[r].lazy = 0;
 72     }
 73 }
 74 void Build(int i,int l,int r)
 75 {
 76     segTree[i].l = l;
 77     segTree[i].r = r;
 78     segTree[i].val = -1;
 79     segTree[i].lazy = 0;
 80     if(l == r)return;
 81     int mid = (l+r)/2;
 82     Build(i<<1,l,mid);
 83     Build((i<<1)|1,mid+1,r);
 84 }
 85 void update(int i,int l,int r,int v)
 86 {
 87     if(segTree[i].l == l && segTree[i].r == r)
 88     {
 89         Update_Same(i,v);
 90         return;
 91     }
 92     push_down(i);
 93     int mid = (segTree[i].l + segTree[i].r)/2;
 94     if(r <= mid)update(i<<1,l,r,v);
 95     else if(l > mid)update((i<<1)|1,l,r,v);
 96     else
 97     {
 98         update(i<<1,l,mid,v);
 99         update((i<<1)|1,mid+1,r,v);
100     }
101 }
102 int query(int i,int u)
103 {
104     if(segTree[i].l == u && segTree[i].r == u)
105         return segTree[i].val;
106     push_down(i);
107     int mid = (segTree[i].l + segTree[i].r)/2;
108     if(u <= mid)return query(i<<1,u);
109     else return query((i<<1)|1,u);
110 }
111 bool used[MAXN];
112 int main()
113 {
114     //freopen("in.txt","r",stdin);
115     //freopen("out.txt","w",stdout);
116     int n;
117     int T;
118     scanf("%d",&T);
119     int iCase = 0;
120     while(T--)
121     {
122         iCase++;
123         printf("Case #%d:\n",iCase);
124         int u,v;
125         memset(used,false,sizeof(used));
126         init();
127         scanf("%d",&n);
128         for(int i = 1;i < n;i++)
129         {
130             scanf("%d%d",&u,&v);
131             used[u] = true;
132             addedge(v,u);
133         }
134         for(int i = 1;i <= n;i++)
135             if(!used[i])
136             {
137                 dfs(i);
138                 break;
139             }
140         Build(1,1,cnt);
141         char op[10];
142         int m;
143         scanf("%d",&m);
144         while(m--)
145         {
146             scanf("%s",op);
147             if(op[0] == 'C')
148             {
149                 scanf("%d",&u);
150                 printf("%d\n",query(1,start[u]));
151             }
152             else
153             {
154                 scanf("%d%d",&u,&v);
155                 update(1,start[u],end[u],v);
156             }
157         }
158     }
159     return 0;
160 }

 

 

 

 

posted on 2013-11-17 22:24  kuangbin  阅读(3560)  评论(0)  编辑  收藏  举报

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