hdoj 3974 Assign the task 【DFS + 线段树区间修改】

Assign the task

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1457    Accepted Submission(s): 667

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

 

题意：给你一棵有N个节点的树，有M次操作。

C x y 表示把x点权值修改为y，该效应会影响到x的所有孩子、孙子节点。Q x 查询x的权值。

思路：求出DFS序，建立一棵线段树，找到每个节点第一次出现的位置left和最后一次出现的位置right，更新时更新区间[left, right]，查询时单点查询即可。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <vector>#include <string>#define INF 0x3f3f3f3f#define eps 1e-8#define MAXN (100000+10)#define MAXM (500000)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%.2lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1#define PI acos(-1.0)using namespace std;struct Tree{    int l, r, len;    int lazy, sum;};Tree tree[MAXN<<2];void PushUp(int o){    tree[o].sum = tree[ll].sum + tree[rr].sum;}void PushDown(int o){    if(tree[o].lazy != -1)    {        tree[ll].lazy = tree[rr].lazy = tree[o].lazy;        tree[ll].sum = tree[o].lazy * tree[ll].len;        tree[rr].sum = tree[o].lazy * tree[rr].len;        tree[o].lazy = -1;    }}void Build(int o, int l, int r){    tree[o].l = l; tree[o].r = r;    tree[o].len = r-l+1; tree[o].sum = -1;    tree[o].lazy = -1;    if(l == r)        return ;    int mid = (l + r) >> 1;    Build(lson); Build(rson);}void Update(int o, int L, int R, int v){    if(tree[o].l >= L && tree[o].r <= R)    {        tree[o].lazy = v;        tree[o].sum = tree[o].len * v;        return ;    }    PushDown(o);    int mid = (tree[o].l + tree[o].r) >> 1;    if(R <= mid)        Update(ll, L, R, v);    else if(L > mid)        Update(rr, L, R, v);    else    {        Update(ll, L, mid, v);        Update(rr, mid+1, R, v);    }    PushUp(o);}int Query(int o, int pos){    if(tree[o].l == tree[o].r)        return tree[o].sum;    PushDown(o);    int mid = (tree[o].l + tree[o].r) >> 1;    if(pos <= mid)        return Query(ll, pos);    else        return Query(rr, pos);}struct Edge{    int from, to, next;};Edge edge[MAXM];int head[MAXN], edgenum;void init(){    edgenum = 0;    CLR(head, -1);}void addEdge(int u, int v){    Edge E = {u, v, head[u]};    edge[edgenum] = E;    head[u] = edgenum++;}int vs[MAXN], dfs_clock;void DFS(int u, int fa, int d){    vs[dfs_clock++] = u;    for(int i = head[u]; i != -1; i = edge[i].next)    {        int v = edge[i].to;        if(v == fa) continue;        DFS(v, u, d+1);        vs[dfs_clock++] = u;    }}int pre[MAXN];int lp[MAXN], rp[MAXN];int main(){    int t, kcase = 1; Ri(t);    W(t)    {        int N, Q; Ri(N);        init();        for(int i = 1; i <= N; i++)            pre[i] = i;        for(int i = 0; i < N-1; i++)        {            int u, v;            Ri(u); Ri(v);            pre[u] = v;            addEdge(u, v);            addEdge(v, u);        }        int root;        for(int i = 1; i <= N; i++)        {            if(pre[i] == i)            {                root = i;                break;            }        }        dfs_clock = 1; DFS(root, -1, 1);        CLR(lp, 0); CLR(rp, 0);        for(int i = 1; i < dfs_clock; i++)        {            printf("%d ", vs[i]);            if(lp[vs[i]] == 0)                lp[vs[i]] = rp[vs[i]] = i;            else                rp[vs[i]] = i;        }        printf("\n");        Ri(Q); Build(1, 1, dfs_clock-1);        printf("Case #%d:\n", kcase++);        W(Q)        {            char op[2];            Rs(op); int x, y;            if(op[0] == 'T')            {                Ri(x); Ri(y);                Update(1, lp[x], rp[x], y);            }            else            {                Ri(x);                Pi(Query(1, lp[x]));            }        }    }    return 0;}