传送门:CodeForces-960E
思路分析
考虑进行树DP,显然偶数路径的总权值是0,因为来回抵消了,所以统计答案的时候只需要计算奇数路径
$siz[u][0/1]$记录u的子树中,以u结尾的偶/奇路径个数
$dis[u][0/1]$记录u的子树中,以u结尾的偶/奇路径权值和
最后答案*2,因为双向都要算一遍,再加上单个点的贡献
AC代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
bool IOerror=0;
inline char nc() {
static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
if(p1==pend) {
p1=buf;
pend=buf+fread(buf,1,BUF_SIZE,stdin);
if(pend==p1) {
IOerror=1;
return -1;
}
}
return *p1++;
}
inline bool blank(char ch) {
return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
}
template<class T> inline bool read(T &x) {
bool sign=0;
char ch=nc();
x=0;
for(; blank(ch); ch=nc());
if(IOerror)return false;
if(ch=='-')sign=1,ch=nc();
for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
if(sign)x=-x;
return true;
}
template<class T,class... U>bool read(T& h,U&... t) {
return read(h)&&read(t...);
}
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int INF = 0x3f3f3f3f;
const int N = 2e5+10;
const int mod= 1e9+7;
#define int long long
vector<int>e[N];
int dis[N][2],siz[N][2],n,ans,w[N];
void dfs(int u,int fa){
siz[u][1]=1;
dis[u][1]=(w[u]+mod)%mod;
for(auto v:e[u]){
if(v==fa) continue;
dfs(v,u);
ans=(ans+dis[u][0]*siz[v][1]%mod+dis[v][1]*siz[u][0]%mod)%mod;
ans=(ans+dis[u][1]*siz[v][0]%mod+dis[v][0]*siz[u][1]%mod)%mod;
siz[u][0]+=siz[v][1];
siz[u][1]+=siz[v][0];
dis[u][0]=(dis[u][0]+dis[v][1]-w[u]*siz[v][1]%mod+mod)%mod;
dis[u][1]=(dis[u][1]+dis[v][0]+w[u]*siz[v][0]%mod+mod)%mod;
}
}
signed main() {
#ifdef xiaofan
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
#endif
read(n);
for(int i=1;i<=n;i++) read(w[i]);
for(int i=1;i<n;i++){
int u,v;
read(u,v);
e[u].pb(v);
e[v].pb(u);
}
dfs(1,1);
ans*=2;
ans%=mod;
for(int i=1;i<=n;i++) ans=(ans+w[i]+mod)%mod;
cout<<ans<<endl;
return 0;
}
?