【NOIP模拟题】T3 【递归】【复杂公式】【模拟】

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<set>
#include<map>
#include<queue>
#include<algorithm>
#include<vector>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<stack>
#define INF 2100000000
#define ll long long
#define clr(x)  memset(x,0,sizeof(x))
#define clrmax(x)  memset(x,127,sizeof(x))

using namespace std;

inline ll read()
{
char c;
ll ret=0;
while(!(c>='0'&&c<='9'))
c=getchar();
while(c>='0'&&c<='9')
{
ret=(c-'0')+(ret<<1)+(ret<<3);
c=getchar();
}
return ret;
}

#define M 65
#define P 1000000007
#define p(x,y) make_pair((x),(y))
ll n,num[M],a[M],b[M],c[M],d[M],l[M];
ll ans[M];
map<ll,map<ll,ll> >dis;//map[i][j]树i上其余所有节点到j节点的距离和
map<ll,map<pair<ll,ll>,ll> >len;//树i上两点之间距离

inline void get_len(int x,ll p1,ll p2)
{
ll ret=0;
if(p1>p2)swap(p1,p2);
if(x==0||p1==p2||len[x][p(p1,p2)])return ;
if(p1+1<=num[a[x]]&&p2+1<=num[a[x]])
{
get_len(a[x],p1,p2);
len[x][p(p1,p2)]=len[a[x]][p(p1,p2)];
len[x][p(p2,p1)]=len[a[x]][p(p1,p2)];
return ;
}
if(p1+1>num[a[x]]&&p2+1>num[a[x]])
{
get_len(b[x],p1-num[a[x]],p2-num[a[x]]);
ret=len[b[x]][p(p1-num[a[x]],p2-num[a[x]])];
len[x][p(p1,p2)]=ret;
len[x][p(p2,p1)]=ret;
return ;
}
get_len(a[x],p1,c[x]);
get_len(b[x],p2-num[a[x]],d[x]);
ret=len[a[x]][p(p1,c[x])]+len[b[x]][p(p2-num[a[x]],d[x])]+l[x];
ret%=P;
len[x][p(p1,p2)]=ret;
len[x][p(p2,p1)]=ret;
}

inline void get_num(int x,ll y)
{
if(dis[x][y]||x==0)return ;
ll ret=0;
int B=0;
if(y+1>num[a[x]])
{
B=1;
y-=num[a[x]];
swap(a[x],b[x]);
swap(c[x],d[x]);
}
get_num(a[x],y);
get_num(b[x],d[x]);
get_len(a[x],y,c[x]);
ret+=dis[b[x]][d[x]];
ret+=l[x]*num[b[x]]%P;
ret+=len[a[x]][p(y,c[x])]*num[b[x]]%P;
ret+=dis[a[x]][y];
ret%=P;
if(B)
{
swap(a[x],b[x]);
swap(c[x],d[x]);
y+=num[a[x]];
}
dis[x][y]=ret;
}

void work(int i)
{
a[i]=read();b[i]=read();c[i]=read();d[i]=read();l[i]=read();
get_num(a[i],c[i]);
get_num(b[i],d[i]);
num[i]=num[a[i]]+num[b[i]];
ans[i]=ans[a[i]]+ans[b[i]]+l[i]%P*num[a[i]]%P*num[b[i]]%P;
int t1=dis[a[i]][c[i]],t2=dis[b[i]][d[i]];
ans[i]+=t1*num[b[i]]%P+t2*num[a[i]]%P;
ans[i]%=P;
}

int main()
{
freopen("zi.in","r",stdin);
freopen("zi.out","w",stdout);
n=read();num[0]=1;
for(int i=1;i<=n;i++)
work(i);
for(int i=1;i<=n;i++)
cout<<ans[i]<<'\n';
return 0;
}