2021年度训练联盟热身训练赛第一场 I. Full Depth Morning Show
作者:互联网
链接
https://ac.nowcoder.com/acm/contest/12606/I
题意
树上对于任意节点 u u u 求 ∑ v = 1 n [ d u , v ∗ ( t u + t v ) ] \sum_{v=1}^{n}{[d_{u,v}*(t_u+t_v)]} ∑v=1n[du,v∗(tu+tv)]
思路
换根 DP
记 r e s u = ∑ v = 1 n [ d u , v ∗ ( t u + t v ) ] res_u=\sum_{v=1}^{n}{[d_{u,v}*(t_u+t_v)]} resu=∑v=1n[du,v∗(tu+tv)]
r e s u = ∑ v = 1 n [ d u , v ∗ ( t u + t v ) ] = t u ∗ ∑ v = 1 n d u , v + ∑ v = 1 n ( d u , v ∗ t v ) res_{u}=\sum_{v=1}^{n}{[d_{u,v}*(t_u+t_v)}]=t_u*\sum_{v=1}^{n}{d_{u,v}}+\sum_{v=1}^{n}{(d_{u,v}*t_v)} resu=∑v=1n[du,v∗(tu+tv)]=tu∗∑v=1ndu,v+∑v=1n(du,v∗tv)
记 a u = ∑ v = 1 n d u , v a_u=\sum_{v=1}^{n}{d_{u,v}} au=∑v=1ndu,v, b u = ∑ v = 1 n ( d u , v ∗ t v ) b_u=\sum_{v=1}^{n}{(d_{u,v}*t_v)} bu=∑v=1n(du,v∗tv), s z u sz_u szu 为节点 u u u 的子树节点数, s u m u sum_u sumu 为节点 u u u 的子树节点权值和, u ′ u' u′ 为节点 u u u 的儿子节点, S U M = ∑ i = 1 n t i SUM=\sum_{i=1}^{n}{t_i} SUM=∑i=1nti
a u ′ = a u − d u , u ′ ∗ s z u + d u , u ′ ∗ ( n − s z u ′ ) = a u + d u , u ′ ∗ ( n − 2 ∗ s z u ′ ) a_{u'}=a_u-d_{u,u'}*sz_u+d_{u,u'}*(n-sz_{u'})=a_u+d_{u,u'}*(n-2*sz_{u'}) au′=au−du,u′∗szu+du,u′∗(n−szu′)=au+du,u′∗(n−2∗szu′)
b u ′ = b u − d u , u ′ ∗ s u m u + d u , u ′ ∗ ( S U M − s u m u ′ ) = b u + d u , u ′ ∗ ( S U M − 2 ∗ s u m u ′ ) b_{u'}=b_u-d_{u,u'}*sum_u+d_{u,u'}*(SUM-sum_{u'})=b_u+d_{u,u'}*(SUM-2*sum_{u'}) bu′=bu−du,u′∗sumu+du,u′∗(SUM−sumu′)=bu+du,u′∗(SUM−2∗sumu′)
r e s u ′ = t u ′ ∗ a u ′ + b u ′ res_{u'}=t_{u'}*a_{u'}+b_{u'} resu′=tu′∗au′+bu′
两次 DFS, O ( n ) O(n) O(n) 求解
代码
#include <bits/stdc++.h>
#define SZ(x) (int)(x).size()
#define ALL(x) (x).begin(),(x).end()
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define FI first
#define SE second
using namespace std;
typedef double DB;
typedef long double LD;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef vector<PII> VPII;
// head
const int N=1e5+5;
int n;
LL t[N],d[N],sz[N],sum[N],a[N],b[N],res[N];
VPII g[N];
void dfs1(int u,int fa) {
sz[u]=1;
sum[u]=t[u];
for(auto x:g[u]) {
int v=x.FI,w=x.SE;
if(v==fa) continue;
d[v]=d[u]+w;
a[1]+=d[v];
b[1]+=d[v]*t[v];
dfs1(v,u);
sz[u]+=sz[v];
sum[u]+=sum[v];
}
}
void dfs2(int u,int fa) {
for(auto x:g[u]) {
int v=x.FI,w=x.SE;
if(v==fa) continue;
a[v]=a[u]+w*(n-2*sz[v]);
b[v]=b[u]+w*(sum[1]-2*sum[v]);
res[v]=a[v]*t[v]+b[v];
dfs2(v,u);
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cin>>n;
for(int i=1;i<=n;i++) cin>>t[i];
for(int i=1;i<n;i++) {
int u,v,w;
cin>>u>>v>>w;
g[u].EB(v,w);
g[v].EB(u,w);
}
dfs1(1,0);
dfs2(1,0);
res[1]=a[1]*t[1]+b[1];
for(int i=1;i<=n;i++) cout<<res[i]<<'\n';
return 0;
}
标签:sz,typedef,Full,Show,int,sum,Morning,du,1n 来源: https://blog.csdn.net/qq_43651414/article/details/114646453