HDU3974 Assign the task —— dfs时间戳 + 线段树

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mob604756ef7d06 2017-10-24 21:55:00

文章标签 #include 结点时间戳 i++ 子树 文章分类 后端开发

题目链接：https://vjudge.net/problem/HDU-3974

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.

InputThe 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)OutputFor 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

Sample Output

Case #1:
-1 
1 
2

题解：

1.可知：对一棵树进行dfs（前序遍历），并为每个结点分配一个时间戳，表明该结点是第几个被访问的结点。对于某一个结点（非叶子），它的所有子孙的遍历次序是紧跟着当前节点的。

2.根据时间戳dfn[u]，把每个结点u映射到一维数组上。设le[u]为子树u开始访问的时间戳，可知le[u]=dfn[u]；ri[u]为子树u结束访问的时间戳，可知ri[u]为结点u最后被访问的子孙的时间戳。所以结点u的作用域就是： [ le[u], ri[u] ]。因此，我们就可以用线段树的区间修改来进行维护了。

学习之处：

根据DFS的特性：对一棵树进行先序遍历，对于当前结点u，它的所有子孙的遍历次序是紧跟着当前节点的。

因此，我们可以根据时间戳，把一棵树映射到到一维数组上，且可以知道每棵子树的确切位置。

代码如下：

HDU3974 Assign the task —— dfs时间戳 + 线段树_时间戳

HDU3974 Assign the task —— dfs时间戳 + 线段树_i++_02

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cmath>
  5 #include <algorithm>
  6 #include <vector>
  7 #include <queue>
  8 #include <stack>
  9 #include <map>
 10 #include <string>
 11 #include <set>
 12 using namespace std;
 13 typedef long long LL;
 14 const double EPS = 1e-8;
 15 const int INF = 2e9;
 16 const LL LNF = 2e18;
 17 const int MAXN = 5e4+10;
 18 
 19 vector<int>g[MAXN];
 20 int have_fa[MAXN];
 21 int index, dfn[MAXN], le[MAXN], ri[MAXN];
 22 int task[MAXN<<2];
 23 
 24 void dfs(int u)
 25 {
 26     dfn[u] = le[u] = ++index;
 27     for(int i = 0; i<g[u].size(); i++)
 28         dfs(g[u][i]);
 29     ri[u] = index;
 30 }
 31 
 32 void push_down(int u)
 33 {
 34     if(task[u]!=-1)
 35     {
 36         task[u*2] = task[u*2+1] = task[u];
 37         task[u] = -1;
 38     }
 39 }
 40 
 41 void set_val(int u, int l, int r, int x, int y, int val)
 42 {
 43     if(x<=l && r<=y)
 44     {
 45         task[u] = val;
 46         return;
 47     }
 48 
 49     push_down(u);
 50     int mid = (l+r)>>1;
 51     if(x<=mid) set_val(u*2, l, mid, x, y, val);
 52     if(y>=mid+1) set_val(u*2+1, mid+1, r, x, y, val);
 53 }
 54 
 55 int query(int u, int l, int r, int x)
 56 {
 57     if(l==r) return task[u];
 58 
 59     push_down(u);
 60     int mid = (l+r)>>1;
 61     if(x<=mid) return query(u*2, l, mid, x);
 62     else return query(u*2+1, mid+1, r, x);
 63 }
 64 
 65 int main()
 66 {
 67     int n, m, T;
 68     scanf("%d", &T);
 69     for(int kase = 1; kase<=T; kase++)
 70     {
 71         scanf("%d", &n);
 72         for(int i = 1; i<=n; i++)
 73             g[i].clear(), have_fa[i] = 0;
 74         for(int i = 1; i<n; i++)
 75         {
 76             int u, v;
 77             scanf("%d%d", &u, &v);
 78             g[v].push_back(u);
 79             have_fa[u] = 1;
 80         }
 81 
 82         index = 0;
 83         for(int i = 1; i<=n; i++)
 84             if(!have_fa[i])
 85                 dfs(i);
 86 
 87         scanf("%d", &m);
 88         memset(task, -1, sizeof(task));
 89         printf("Case #%d:\n", kase);
 90         for(int i = 1; i<=m; i++)
 91         {
 92             char op[2]; int x, y;
 93             scanf("%s", op);
 94             if(op[0]=='T')
 95             {
 96                 scanf("%d%d", &x, &y);
 97                 set_val(1, 1, n, le[x], ri[x], y);
 98             }
 99             else
100             {
101                 scanf("%d", &x);
102                 printf("%d\n", query(1, 1, n, dfn[x]));
103             }
104         }
105     }
106 }