HDU3887 Counting Offspring [2017年6月计划 树上问题03]
2024-09-24 16:59:20
Counting Offspring
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2809 Accepted Submission(s): 981
Problem Description
You
are given a tree, it’s root is p, and the node is numbered from 1 to n.
Now define f(i) as the number of nodes whose number is less than i in
all the succeeding nodes of node i. Now we need to calculate f(i) for
any possible i.
are given a tree, it’s root is p, and the node is numbered from 1 to n.
Now define f(i) as the number of nodes whose number is less than i in
all the succeeding nodes of node i. Now we need to calculate f(i) for
any possible i.
Input
Multiple cases (no more than 10), for each case:
The first line contains two integers n (0<n<=10^5) and p, representing this tree has n nodes, its root is p.
Following n-1 lines, each line has two integers, representing an edge in this tree.
The input terminates with two zeros.
The first line contains two integers n (0<n<=10^5) and p, representing this tree has n nodes, its root is p.
Following n-1 lines, each line has two integers, representing an edge in this tree.
The input terminates with two zeros.
Output
For each test case, output n integer in one line representing f(1), f(2) … f(n), separated by a space.
Sample Input
15 7
7 10
7 1
7 9
7 3
7 4
10 14
14 2
14 13
9 11
9 6
6 5
6 8
3 15
3 12
0 0
7 10
7 1
7 9
7 3
7 4
10 14
14 2
14 13
9 11
9 6
6 5
6 8
3 15
3 12
0 0
Sample Output
0 0 0 0 0 1 6 0 3 1 0 0 0 2 0
Author
bnugong
Source
Recommend
//====================================
被格式错误卡了半个小时
开始多了空格,去了空格,不对
后来发现多组数据要加空行,然后在数据之间加了空行,不对
看了看题解才发现末尾要多一个空行
竟无语凝噎
//====================================
跟树状数组求逆序对的思想类似,大家可以去看那一道题的思路
#include <bits/stdc++.h> inline void read(int &x)
{
char ch = getchar();char c = ch;x = 0;
while(ch < '0' || ch > '9')c = ch, ch = getchar();
while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
if(c == '-')x = -x;
}
inline int lowbit(int &a){return a & (-a);}
const int MAXN = 500000 + 10; int n,root,tmp1,tmp2; struct Edge{int u,v,next;}edge[MAXN << 1];
int head[MAXN], cnt, l[MAXN << 1], r[MAXN << 1], bit[MAXN << 1];
inline void insert(int a, int b){edge[++cnt] = Edge{a,b,head[a]};head[a] = cnt;}
int b[MAXN], stack[MAXN], top, rank; void dfs(int root)
{
register int u,v,pos;
stack[++top] = root;
b[root] = 1;
while(top)
{
u = stack[top--];
if(l[u])
{
r[u] = ++rank;
continue;
}
stack[++top] = u;
l[u] = ++rank;
for(pos = head[u];pos;pos = edge[pos].next)
{
v = edge[pos].v;
if(b[v])continue;
b[v] = true;
stack[++top] = v;
}
}
} inline void modify(int p, int k)
{
register int tmp = n << 1;
for(;p <= tmp;p += lowbit(p))
bit[p] += k;
} inline int ask(int p)
{
register int ans = 0;
for(;p;p -= lowbit(p))
ans += bit[p];
return ans;
} bool ok;
int main()
{
while(true)
{
read(n);read(root);
if(!(n || root))break;
memset(edge, 0, sizeof(edge));
memset(head, 0, sizeof(head));
memset(l, 0, sizeof(l));
memset(r, 0, sizeof(r));
cnt = 0;
memset(bit, 0, sizeof(bit));
memset(b, 0, sizeof(b));
memset(stack, 0, sizeof(stack));
top = 0;
rank = 0;
register int i;
for(i = 1;i < n;++ i)
{
read(tmp1);read(tmp2);
insert(tmp1, tmp2);
insert(tmp2, tmp1);
}
dfs(root);
printf("%d", ask(r[1]) - ask(l[1] - 1));
modify(l[1], 1);
for(i = 2;i <= n;i ++)
{
printf(" %d", ask(r[i]) - ask(l[i] - 1));
modify(l[i], 1);
}
printf("\n");
}
return 0;
}
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