牛客练习赛6 B 点权和 树点权和

https://www.nowcoder.com/acm/contest/26/B

考虑每次修改的贡献

 

考虑这次操作x造成的影响：

x对自己的贡献：

x的节点值会加上deg[x]+1，deg为度数

x对上方的贡献：

x的父亲节点的值会加上2

x的父亲的父亲节点的值会加上1

x的父亲节点的儿子值会加上1

x对下方的贡献：

x的儿子节点值会加上2

x的儿子的儿子节点值会加上1

///                 .-~~~~~~~~~-._       _.-~~~~~~~~~-.///             __.'              ~.   .~              `.__///           .'//                  \./                  \\`.///        .'//                     |                     \\`.///       .'// .-~"""""""~~~~-._     |     _,-~~~~"""""""~-. \\`.///     .'//.-"                 `-.  |  .-'                 "-.\\`.///   .'//______.============-..   \ | /   ..-============.______\\`./// .'______________________________\|/______________________________`.#pragma GCC optimize(2)#pragma comment(linker, "/STACK:102400000,102400000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define s_1(x) scanf("%d",&x)#define s_2(x,y) scanf("%d%d",&x,&y)#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)#define S_1(x) scan_d(x)#define S_2(x,y) scan_d(x),scan_d(y)#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define fOR(n,x,i) for(int i=n;i>=x;i--)#define fOr(n,x,i) for(int i=n;i>x;i--)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double#define ll long long#define mp make_pair#define pb push_backtypedef long long LL;typedef pair <int, int> ii;const int INF=~0U>>1;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=1e5+10;const int maxx=1e6+10;const double EPS=1e-8;const double eps=1e-8;const int mod=19260817;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );//cerr << "run time is " << clock() << endl;//void readString(string &s)//{//static char str[maxn];//scanf("%s", str);//s = str;//}int n,m;int dep[maxn],fa[maxn];LL tag[maxn];//tag[x] x被操作的次数 LL nowtag[maxn],sontag[maxn];//nowtag[x]从子树上贡献上来的次数//sontag[x]从x的儿子被操作的次数 void solve(){s_2(n,m);FOr(1,n,i){ int x;s_1(x);fa[i+1]=x;dep[x]++;dep[i+1]++;}LL Hash=0;fa[1]=0;FOR(1,m,i){int x;s_1(x);tag[x]++;nowtag[x]+=dep[x]+1;nowtag[fa[x]]+=2;nowtag[fa[fa[x]]]++;sontag[fa[x]]++;LL ans=nowtag[x];ans=(ans+tag[fa[x]]*2ll+mod)%mod;//x的父亲每次操作对x有2的贡献 ans=(ans+tag[fa[fa[x]]]+mod)%mod;//父亲的父亲每次操作对x有1的贡献 ans=(ans+sontag[fa[x]]-tag[x]+mod)%mod;//x的父亲的其他儿子对于x的贡献 Hash=(ans*i%mod+Hash+mod)%mod;//cout<<ans<<endl;}print(Hash);}int main(){    //freopen( "1.in" , "r" , stdin );    //freopen( "1.out" , "w" , stdout );    int t=1;    //init();    //s_1(t);    for(int cas=1;cas<=t;cas++)    {        //printf("Case #%d: ",cas);        solve();    }}