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.

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.

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

Sample Output

0 0 0 0 0 1 6 0 3 1 0 0 0 2 0

Author

bnugong

Source

2011 Multi-University Training Contest 5 - Host by BNU

Recommend

lcy   |   We have carefully selected several similar problems for you:  3450 1166 3030 1541 3743

//====================================

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;
}