bzoj 3727: Final Zadanie 思维题

题目:

Description

吉丽YY了一道神题，题面是这样的：

“一棵n个点的树，每条边长度为1，第i个结点居住着a[i]个人。假设在i结点举行会议，所有人都从原住址沿着最短路径来到i结点，行走的总路程为b[i]。输出所有b[i]。”

吉丽已经造好了数据，但熊孩子把输入文件中所有a[i]给删掉了。你能帮他恢复吗？

题解:

对于节点\(u\)设其父亲为\(fa_u\).子树的\(a_i\)之和为\(sum_i\)

设\(SUM = \sum_{u \in G}a_u\)

则对于任意的\(u \neq 1\)有:\(b_u - b_{fa_u} = SUM - 2*sum_u\)

且\(b_1 = \sum_{u \neq 1}sum_u\)

然后我们将所有的\(b_u - b_{fa_u} = SUM - 2*sum_u\)求和

得:\(\sum (b_u - b_{fa_u}) = (n-1)*SUM - 2\sum_{u \neq 1}sum_u\)

我们将\(b_1 = \sum_{u \neq 1}sum_u\)翻倍加上去有.

\(2*b_1 = \sum (b_u - b_{fa_u}) = (n-1)*SUM\)

于是我们有:\(SUM = \frac{2*b_1 + \sum_{u \neq 1}(b_u - b_{fa_u})}{n-1}\)

既然我们得到了SUM,那么将其代入到所有的\(b_u - b_{fa_u} = SUM - 2*sum_u\)中即可

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
inline void read(int &x){
	x=0;char ch;bool flag = false;
	while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
	while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
const int maxn = 300010;
int n;
struct Egde{
	int to,next;
}G[maxn<<1];
int head[maxn],cnt;
void add(int u,int v){
	G[++cnt].to = v;
	G[cnt].next = head[u];
	head[u] = cnt;
}
int b[maxn],sum[maxn],fa[maxn],a[maxn];
#define v G[i].to
void dfs1(int u){
	for(int i = head[u];i;i=G[i].next){
		if(v == fa[u]) continue;
		fa[v] = u;dfs1(v);
	}
}
void dfs2(int u){
	a[u] = sum[u];
	for(int i = head[u];i;i=G[i].next){
		if(v == fa[u]) continue;
		dfs2(v);a[u] -= sum[v];
	}
}
#undef v
inline void init(){
	memset(head,0,sizeof head);
	cnt = 0;
}
int main(){
	init();read(n);
	for(int i=1,u,v;i<n;++i){
		read(u);read(v);
		add(u,v);add(v,u);
	}
	memset(sum,0,sizeof sum);
	memset(fa,0,sizeof fa);
	for(int i=1;i<=n;++i) read(b[i]);
	dfs1(1);
	ll x = 0;
	for(int i=2;i<=n;++i){
		x += b[i] - b[fa[i]];
	}x += 2LL*b[1];
	ll SUM = x/(n-1);
	sum[1] = SUM;
	for(int i=2;i<=n;++i){
		sum[i] = SUM - (b[i] - b[fa[i]]);
		sum[i] >>= 1;
	}
	dfs2(1);
	for(int i=1;i<=n;++i){
		printf("%d",a[i]);
		if(i != n) putchar(' ');
		else putchar('\n');
	}
	return 0;
}