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2017 ACM-ICPC 亚洲区(西安赛区)网络赛 G. Xor

时间:2017-09-16 21:58:21      阅读:398      评论:0      收藏:0      [点我收藏+]

标签:html   long   tac   with   node   sub   lis   const   htm   

There is a tree with nn nodes. For each node, there is an integer value a_ia?i??, (1 \le a_i \le 1,000,000,0001a?i??1,000,000,000 for 1 \le i \le n1in). There is qq queries which are described as follow: Assume the value on the path from node aa to node bb is t_0, t_1, \cdots t_mt?0??,t?1??,?t?m??. You are supposed to calculate t_0t?0?? xor t_kt?k?? xor t_{2k}t?2k?? xor ... xor t_{pk}t?pk?? (pk \le m)(pkm).

Input Format

There are multi datasets. (\sum n \le 50,000, \sum q \le 500,000)(n50,000,q500,000).

For each dataset: In the first n-1n1 lines, there are two integers u,vu,v, indicates there is an edge connect node uu and node vv.

In the next nn lines, There is an integer a_ia?i?? (1 \le a_i \le 1,000,000,0001a?i??1,000,000,000).

In the next qq lines, There is three integers a,ba,band kk. (1 \le a,b,k \le n1a,b,kn).

Output Format

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

样例输入

5 6
1 5
4 1
2 1
3 2
19
26
0
8
17
5 5 1
1 3 2
3 2 1
5 4 2
3 4 4
1 4 5

样例输出

17
19
26
25
0
19
分析:求树上路径从距离起点为k的倍数的权值和;
   大于根号n暴力,小于的话预处理,处理到根的前缀异或和;

代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <bitset>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <cassert>
#include <ctime>
#define rep(i,m,n) for(i=m;i<=(int)n;i++)
#define inf 0x3f3f3f3f
#define mod 1000000007
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define sys system("pause")
#define ls (rt<<1)
#define rs (rt<<1|1)
#define all(x) x.begin(),x.end()
const int maxn=5e4+10;
const int N=2e5+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qmul(ll p,ll q,ll mo){ll f=0;while(q){if(q&1)f=(f+p)%mo;p=(p+p)%mo;q>>=1;}return f;}
ll qpow(ll p,ll q,ll mo){ll f=1;while(q){if(q&1)f=f*p%mo;p=p*p%mo;q>>=1;}return f;}
int n,m,k,t,dp[maxn][310],fa[20][maxn],dep[maxn],a[maxn],sz,q,head[maxn],tot;
struct node
{
    int to,nxt;
}e[maxn<<1];
void add(int x,int y)
{
    e[tot].to=y;
    e[tot].nxt=head[x];
    head[x]=tot++;
}
int lca(int x,int y)
{
    int i;
    if(dep[x]<dep[y])swap(x,y);
    for(i=19;i>=0;i--)if(dep[fa[i][x]]>=dep[y])x=fa[i][x];
    if(x==y)return x;
    for(i=19;i>=0;i--)
    {
        if(fa[i][x]!=fa[i][y])
        {
            x=fa[i][x],
            y=fa[i][y];
        }
    }
    return fa[0][x];
}
int find(int x,int y)
{
    int i;
    for(i=19;i>=0;i--)
    {
        if(y>>i&1)
        {
            x=fa[i][x];
            if(x==0)return 0;
        }
    }
    return x;
}
void dfs(int x,int y)
{
    int i;
    dep[x]=dep[y]+1;
    for(i=1;fa[i-1][fa[i-1][x]];i++)
    {
        fa[i][x]=fa[i-1][fa[i-1][x]];
    }
    rep(i,1,sz)
    {
        dp[x][i]=a[x];
        dp[x][i]^=dp[find(x,i)][i];
    }
    for(i=head[x];i!=-1;i=e[i].nxt)
    {
        int z=e[i].to;
        if(z==y)continue;
        fa[0][z]=x;
        dfs(z,x);
    }
}
int main(){
    int i,j;
    while(~scanf("%d%d",&n,&q))
    {
        sz=round(sqrt(n));
        rep(i,1,n)
        {
            head[i]=-1;
            rep(j,0,19)fa[j][i]=0;
        }
        tot=0;
        rep(i,1,n-1)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            add(x,y);add(y,x);
        }
        rep(i,1,n)scanf("%d",&a[i]);
        dfs(1,0);
        while(q--)
        {
            int x,y,k;
            int ret=0;
            scanf("%d%d%d",&x,&y,&k);
            if(x==y)
            {
                printf("%d\n",a[x]);
                continue;
            }
            int fa=lca(x,y),len=dep[x]+dep[y]-2*dep[fa],pos;
            if((dep[x]-dep[fa])%k==0&&fa!=x&&fa!=y)ret^=a[fa];
            if(k>sz)
            {
                if(fa!=x)
                {
                    pos=x;
                    ret^=a[pos];
                    while(dep[j=find(pos,k)]>=dep[fa])
                    {
                        pos=j;
                        ret^=a[pos];
                    }
                }
                if(fa!=y)
                {
                    pos=find(y,len%k);
                    if(dep[pos]>=dep[fa])
                    {
                        ret^=a[pos];
                        while(dep[j=find(pos,k)]>=dep[fa])
                        {
                            pos=j;
                            ret^=a[pos];
                        }
                    }
                }
            }
            else
            {
                int len1=dep[x]-dep[fa];
                if(fa!=x)
                {
                    len1=len1/k*k;
                    pos=find(x,len1);
                    ret=(ret^dp[x][k]^dp[pos][k]^a[pos]);
                }
                if(fa!=y)
                {
                    int st=find(y,len%k);
                    if(dep[st]>=dep[fa])
                    {
                        len1=dep[st]-dep[fa];
                        len1=len1/k*k;
                        pos=find(st,len1);
                        ret=(ret^dp[st][k]^dp[pos][k]^a[pos]);
                    }
                }
            }
            printf("%d\n",ret);
        }
    }
    return 0;
}
/*
6 1
3 2
1 4
4 6
6 2
1 5
60
2
75
34
60
15
4 6 1
ans:45
*/

2017 ACM-ICPC 亚洲区(西安赛区)网络赛 G. Xor

标签:html   long   tac   with   node   sub   lis   const   htm   

原文地址:http://www.cnblogs.com/dyzll/p/7532844.html

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