HDU3974 Assign the task

Assign the task

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5032    Accepted Submission(s): 1966

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.

 

 

InThe 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

 

思路：

对给定的树，做某个节点下的整点修改。

竟然是用dfs序，作为更新的区间，这实在是太可怕了，虽然除此之外的东西比较简单，我还是wa了不少，是在是太菜了。

得到一个教训，可以push_down的，就一定要及时push_down！

#include<iostream>
#include<vector>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn  = 5e5+8;
int First[maxn<<1],LAST[maxn<<1],wa;
vector<int>u[maxn];
bool book[maxn];
struct node
{
    int l,r;
    int num;
    bool lazy;
}tree[maxn<<2];
void dfs(int t)
{
    int siz=u[t].size();
    First[t]=++wa;
    for(int i=0;i<siz;i++){
        dfs(u[t][i]);
    }
    LAST[t]=++wa;
}

void build(int t,int l,int r)
{
     tree[t].l=l;tree[t].r=r;
     tree[t].num=-1;tree[t].lazy=false;
     if(l==r){return;}

     int mid=(l+r)>>1;

     build(t<<1,l,mid);
     build((t<<1)|1,mid+1,r);
}

void push_down(int t)
{
    tree[t<<1].num=tree[t].num;
    if(tree[t<<1].l!=tree[t<<1].r){tree[t<<1].lazy=true;}
    tree[(t<<1)|1].num=tree[t].num;
    if(tree[(t<<1)|1].l!=tree[(t<<1)|1].r){tree[(t<<1)|1].lazy=true;}
    tree[t].lazy=false;
}

void update(int t,int l,int r,int x)
{
    if(tree[t].lazy)push_down(t);
    if(tree[t].l==l&&tree[t].r==r){
        if(l!=r)tree[t].lazy=true;
        tree[t].num=x;return ;
    }

    int mid=(tree[t].l+tree[t].r)>>1;
    if(r<=mid){update(t<<1,l,r,x);}
    else if(l>mid){update((t<<1)|1,l,r,x);}
    else {
        update((t<<1),l,mid,x);
        update((t<<1)|1,mid+1,r,x);
    }
}

int query(int t,int x)
{
//    cout<<t<<" "<<x<<" "<<tree[t].lazy<<endl;
    if(tree[t].lazy)push_down(t);
    if(tree[t].l==tree[t].r){
        if(tree[t].l!=x){return -1;}
        return tree[t].num;
    }

    int mid=(tree[t].l+tree[t].r)>>1;
    if(x<=mid){return query(t<<1,x);}
    else if(x>mid){return query((t<<1)|1,x);}
}


int main()
{
    int T,n;
    scanf("%d",&T);
    int re=0;
    while(T--){
        re++;
        printf("Case #%d:\n",re);
        wa=0;
        memset(book,0,sizeof(book));
        memset(First,0,sizeof(First));
        memset(LAST,0,sizeof(LAST));
        scanf("%d",&n);
        for(int i=0;i<=n+1;i++){
            u[i].clear();
        }
        int x,y;
        for(int i=1;i<n;i++){
            scanf("%d%d",&x,&y);
            u[y].push_back(x);
            book[x]++;
        }
        for(int i=1;i<=n;i++){
            if(!book[i]){dfs(i);}
        }


        build(1,1,n*2);


        scanf("%d",&n);
        char p[5];
        for(int i=1;i<=n;i++){
            getchar();
            scanf("%s",p);
            if(p[0]=='C'){
                scanf("%d",&x);
                printf("%d\n",query(1,First[x]));
            }
            else if(p[0]=='T'){
                scanf("%d%d",&x,&y);
                update(1,First[x],LAST[x],y);
            }

        }
    }
    return 0;
}