HDU 3974 Assign the task（dfs序 + 线段树区间赋值，单点查询）

题意：

给一棵树， C x 操作 查询x 节点的权值， T X Y， 将X 这课子树上所有节点权值都赋值为Y。

思路：

因为是子树所有节点。

直接dfs序跑一遍， 将所有节点变成连续的， 然后就可以用线段树了。

剩下的就是  线段树区间赋值，单点查询问题了。

#include <cstdio>#include <cstring>#include <algorithm>#include <vector>using namespace std;const int maxn = 50000 + 10;int T;vector<int>g[maxn];int fa[maxn];int id[maxn];int r[maxn];int fh[maxn];int cnt;void dfs(int cur){    id[cur] = ++cnt;    fh[cnt] = cur;    for (int i = 0; i < g[cur].size(); ++i){        int v = g[cur][i];        dfs(v);    }    r[cur] = cnt;}int ks;int setv[maxn<<2];void pushdown(int o){    if (setv[o] != -1){        setv[o<<1] = setv[o<<1|1] = setv[o];        setv[o] = -1;    }}void update(int L,int R,int c,int l,int r,int o){    if (L <= l && r <= R){        setv[o] = c;        return;    }    int m = l + r >> 1;    pushdown(o);    if (m >= L){        update(L,R,c, l, m, o << 1);    }    if (m < R){        update(L,R,c,m+1,r,o<<1|1);    }}int query(int pos,int l,int r,int o){    if (l == r){        return setv[o];    }    pushdown(o);    int m = l + r >> 1;    if (m >= pos){        return query(pos, l, m, o<<1);    }    else {        return query(pos, m+1,r,o<<1|1);    }}int main(){    scanf("%d",&T);    while(T--){        cnt = 0;        int n;        scanf("%d",&n);        memset(setv,-1,sizeof setv);        memset(fa,-1,sizeof fa);        for (int i = 1; i <= n; ++i) g[i].clear();        int root = 1;        for (int i = 1; i < n; ++i){            int u, v;            scanf("%d %d",&u, &v);            g[v].push_back(u);            fa[u] = v;        }        for (int i = 1; i <= n; ++i){            if (fa[i] == -1) {                root = i;                break;            }        }        dfs(root);        int q;        printf("Case #%d:\n", ++ks);        scanf("%d",&q);        char cmd[3];        int x,y;        while(q--){            scanf("%s%d", cmd,&x);            if (cmd[0] == 'C'){                printf("%d\n", query(id[x], 1,n,1));            }            else {                scanf("%d",&y);                update(id[x], r[x], y, 1, n, 1);            }        }    }    return 0;}

Assign the task

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2801    Accepted Submission(s): 1172

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

 

Recommend

We have carefully selected several similar problems for you:  3972 3981 3979 3977 3973 

 

Statistic | Submit | Discuss | Note