HDU 3947 Assign the task
http://acm.hdu.edu.cn/showproblem.php?pid=3974
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.
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)
5
4 3
3 2
1 3
5 2
5
C 3
T 2 1
C 3
T 3 2
C 3
-1
1
2
代码:
#include <bits/stdc++.h>
using namespace std; const int maxn = 1e5 + 10;
int T, N;
int head[maxn];
int cnt, tot;
int st[maxn], en[maxn];
bool used[maxn]; struct Edge {
int to, nx;
}edge[maxn]; struct Node {
int val;
int l, r;
int lazy;
}node[maxn * 4]; void init() {
cnt = 0;
tot = 0;
memset(head, -1, sizeof(head));
} void addedge(int u, int v) {
edge[tot].to = v;
edge[tot].nx = head[u];
head[u] = tot ++;
} void dfs(int u) {
cnt ++;
st[u] = cnt;
for(int i = head[u]; i != -1; i = edge[i].nx)
dfs(edge[i].to);
en[u] = cnt;
} void Update_Same (int r, int v) {
if(r) {
node[r].val = v;
node[r].lazy = 1;
}
} void push_down(int r) {
if(node[r].lazy) {
Update_Same(r * 2, node[r].val);
Update_Same(r * 2 + 1, node[r].val);
node[r].lazy = 0;
}
} void Build(int i, int l, int r) {
node[i].l = l;
node[i].r = r;
node[i].val = -1;
node[i].lazy = 0;
if(l == r) return ;
int mid = (l + r) / 2;
Build(i * 2, l, mid);
Build(i * 2 + 1, mid + 1, r);
} void update(int i, int l, int r, int v) {
if(node[i].l == l && node[i].r == r) {
Update_Same(i, v);
return;
}
push_down(i);
int mid = (node[i].l + node[i].r) / 2;
if(r <= mid) update(i * 2, l, r, v);
else if(l > mid) update(i * 2 + 1, l, r, v);
else {
update(i * 2, l, mid, v);
update(i * 2 + 1, mid + 1, r, v);
}
} int query(int i, int u) {
if(node[i].l == u && node[i].r == u)
return node[i].val;
push_down(i);
int mid = (node[i].l + node[i].r) / 2;
if(u <= mid) return query(i * 2, u);
else return query(i * 2 + 1, u);
} int main() {
scanf("%d", &T);
for(int t = 1; t <= T; t ++) {
printf("Case #%d:\n", t);
int u, v;
memset(used, false, sizeof(used));
init();
scanf("%d", &N);
for(int i = 1; i < N; i ++) {
scanf("%d%d", &u, &v);
used[u] = true;
addedge(v, u);
}
for(int i = 1; i <= N; i ++) {
if(!used[i]) {
dfs(i);
break;
}
}
Build(1, 1, cnt);
char op[10];
int M;
scanf("%d", &M);
while(M --) {
scanf("%s", op);
if(op[0] == 'C') {
scanf("%d", &u);
printf("%d\n", query(1, st[u]));
} else {
scanf("%d%d", &u, &v);
update(1, st[u], en[u], v);
}
}
}
return 0;
}
线段树 模板题 链表存图 区间修改单点查询 今天小目标之一把这个搞清楚可以自己写粗来
何生枷锁
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