杭电3974 Assign the task(线段树区间更新点查询)

Assign the task

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 794    Accepted Submission(s): 404

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

/*之前比较困惑的是前向星的存储，将前向星倒着存储，从前边开始检索，编号的时候刚好正向编完。。延迟标记的时候记得将lazy标记往左右子树往下标记。。Time:2015-1-22 16:22*/#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1const int MAX=50000+100;struct Edge{    int next,to;}edge[MAX];int head[MAX],EN;void Add(int u,int v){    edge[EN].to=v;    edge[EN].next=head[u];    head[u]=EN++;}bool vis[MAX];int start[MAX],end[MAX];int k;void Init(){    EN=0,k=0;    memset(vis,0,sizeof(vis));    memset(head,-1,sizeof(head));}void DFS(int u){    start[u]=++k;    for(int i=head[u];~i;i=edge[i].next){        int v=edge[i].to;        DFS(v);    }    end[u]=k;}struct Tree{    int val;    bool lazy;}tree[MAX<<2];void Build(int l,int r,int rt){    tree[rt].val=-1;//初始化为-1    tree[rt].lazy=false;    if(l==r) return;    int mid=(l+r)>>1;    Build(lson);    Build(rson);}void pushDown(int l,int r,int rt){    if(tree[rt].lazy){        tree[rt<<1].lazy=true;        tree[rt<<1].val=tree[rt].val;        tree[rt<<1|1].lazy=true;        tree[rt<<1|1].val=tree[rt].val;        tree[rt].lazy=false;    }}void update(int L,int R,int val,int l,int r,int rt){    if(L==l&&R==r){        tree[rt].lazy=true;        tree[rt].val=val;        return;    }    pushDown(l,r,rt);    int mid=(l+r)>>1;    if(R<=mid) update(L,R,val,lson);    else if(L>mid) update(L,R,val,rson);    else{        update(L,mid,val,lson);        update(mid+1,R,val,rson);    }}int Query(int pos,int l,int r,int rt){    if(l==pos&&r==pos){        return tree[rt].val;    }    pushDown(l,r,rt);    int mid=(l+r)>>1;    if(pos<=mid) return Query(pos,lson);    else return Query(pos,rson);}int main(){    int T,n,m;    int u,v;    int nCase=1;    scanf("%d",&T);    while(T--){        scanf("%d",&n);        Init();        for(int i=1;i<n;i++){            scanf("%d%d",&u,&v);            vis[u]=true;            Add(v,u);//倒着添加边        }        for(int i=1;i<=n;i++){        //边是倒着加的从第一个没标记过的开始可以全部访问完。        //前向星是倒着访问边的。倒着加边刚好从第一个开始访问            if(!vis[i]){                DFS(i);                break;            }        }        Build(1,k,1);        char cmd[10];        printf("Case #%d:\n",nCase++);        scanf("%d",&m);        int u,val;        while(m--){            scanf("%s",cmd);            if(cmd[0]=='C'){                scanf("%d",&u);                printf("%d\n",Query(start[u],1,k,1));            }else{                scanf("%d%d",&u,&val);                update(start[u],end[u],val,1,k,1);            }        }    }return 0;}