西安网络赛 xor

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There is a tree with $n$ nodes. For each node, there is an integer value $a_i$ , ( $\le a_i \le 1,000,000,000$ for $\le i \le n$ ). There is $q$ queries which are described as follow: Assume the value on the path from node $a$ to node $b$ is $t_0, t_1, \cdots t_m$ . You are supposed to calculate $t_0$ xor $t_k$ xor $t_{2k}$ xor ... xor $t_{pk}$ $\le m)$ .

Input Format

There are multi datasets. $(\sum n \le 50,000, \sum q \le 500,000)$ .

For each dataset: In the first $n - 1$ lines, there are two integers $u, v$ , indicates there is an edge connect node $u$ and node $v$ .

In the next $n$ lines, There is an integer $a_i$ ( $\le a_i \le 1,000,000,000$ ).

In the next $q$ lines, There is three integers $a, b$ and $k$ . ( $\le a,b,k \le n$ ).

Output Format

For each query, output an integer in one line, without any additional space.

样例输入

5 61 54 12 13 2192608175 5 11 3 23 2 15 4 23 4 41 4 5

样例输出

17192625019
让你求 u v上的 第0,k,2k,3k 点的异或和
暴力异或 我的代码要加输入挂。。估计写丑了
#include <bits/stdc++.h>using namespace std;const int N = 50010;vector<int> e[N];int vis[N];int dep[N];int du[N];int f[N][20];int a[N];int xr[N];namespace fastIO   {        #define BUF_SIZE 100000        //fread -> read        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';        }        inline void read(int &x)      {            char ch;            while(blank(ch = nc()));            if(IOerror) return;            for(x = ch - '0'; (ch = nc()) >= '0' && ch <= '9'; x = x * 10 + ch - '0');        }        #undef BUF_SIZE    };   using namespace fastIO;  void dfs(int u,int v){    xr[u]=xr[v]^a[u];    for(int i=0;i<e[u].size();i++)    {        if(e[u][i]==v) continue;        dep[e[u][i]]=dep[u]+1;        f[e[u][i]][0]=u;        dfs(e[u][i],u);    }}int n;void ST(){    for(int j=1;(1<<j)<=n;j++)    {        for(int i=1;i<=n;i++)        {            if(f[i][j-1]!=-1)            {                f[i][j]=f[f[i][j-1]][j-1];            }        }    }}int lca(int x,int y){    if(dep[x]<dep[y]) swap(x,y);    int t=dep[x]-dep[y];    for(int i=0;i<=17;i++)        if((1<<i)&t) x=f[x][i];    if(x==y) return x;    for(int i=17;i>=0;i--)    {        if(f[x][i]!=f[y][i])            x=f[x][i],y=f[y][i];    }    if(x==y) return x;    return f[x][0];}int num[30];int main(){    int q;    while(read(n),read(q), !IOerror)    {        memset(f,0,sizeof(f));        for(int i=1;i<=n;i++) e[i].clear();        for(int i=1;i<n;i++)        {            int u,v;            read(u),read(v);            e[u].push_back(v);            e[v].push_back(u);        }                for(int i=1;i<=n;i++)            read(a[i]);        dep[1]=1;        dfs(1,0);        ST();        while(q--)        {            int x,y,k;            read(x),read(y),read(k);            if(k==1)            {                int lc=lca(x,y);                printf("%d\n",xr[x]^xr[y]^a[lc] );                continue;            }            vector<int> rr;            int lc=lca(x,y);            int sum=0;            sum^=a[x];            int kk=k,cnt=0;            while(kk)            {                if(kk%2) num[cnt]=1;                else num[cnt]=0;                kk>>=1;                cnt++;            }            while(1)            {                int ww=x;                int flag=0;                for(int i=0;i<cnt;i++)                    if(num[i])                    ww=f[ww][i];                if(dep[ww]<dep[lc])                     break;                x=ww;                sum^=a[x];            }            int len=dep[x]-dep[lc];            int len1=(dep[y]-dep[lc]-(k-len))%k;            if(dep[y]+len1<=dep[lc]||len1<0)             {                printf("%d\n",sum );                continue;            }            for(int i=0;i<=cnt;i++)            if((1<<i)&len1)               y=f[y][i];            sum^=a[y];            while(1)            {                int ww=y;                int flag=0;                for(int i=0;i<cnt;i++)                {                     if(num[i])                    ww=f[ww][i];                                    }                if(dep[ww]<=dep[lc]) break;                y=ww;                sum^=a[y];            }            printf("%d\n",sum );        }    }}

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