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  • Count on the path

    Count on the path

    Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

    Problem Description
    bobo has a tree, whose vertices are conveniently labeled by 1,2,…,n.

    Let f(a,b) be the minimum of vertices not on the path between vertices a and b.

    There are q queries (ui,vi) for the value of f(ui,vi). Help bobo answer them.
     
    Input
    The input consists of several tests. For each tests:

    The first line contains 2 integers n,q (4≤n≤106,1≤q≤106). Each of the following (n - 1) lines contain 2 integers ai,bi denoting an edge between vertices ai and bi (1≤ai,bi≤n). Each of the following q lines contains 2 integer u′i,v′i (1≤ui,vi≤n).

    The queries are encrypted in the following manner.

    u1=u′1,v1=v′1.
    For i≥2, ui=u′i⊕f(ui - 1,vi - 1),vi=v′i⊕f(ui-1,vi-1).

    Note ⊕ denotes bitwise exclusive-or.

    It is guaranteed that f(a,b) is defined for all a,b.

    The task contains huge inputs. `scanf` in g++ is considered too slow to get accepted. You may (1) submit the solution in c++; or (2) use hand-written input utilities.
     
    Output
    For each tests:

    For each queries, a single number denotes the value.
     
    Sample Input
    4 1 1 2 1 3 1 4 2 3 5 2 1 2 1 3 2 4 2 5 1 2 7 6
     
    Sample Output
    4 3 1
    分析:参考http://www.lai18.com/content/8107004.html;
       学到的东西很多,比如求子树的第二小,在当前根下却不在其中一个儿子的子树的最小值;
       bel[i]表示i在1的哪个儿子下,fa[i]表示i的父亲,son[i]表示i的所有子树里面最小的节点,sec_son[i]对所有i的儿子j的son[j]排序后取第二小的;
       dp[i]表示不是i,却是i的父亲的儿子j下的最小son[j],dp1[i]表示从1走到i所有分支子树下的最小节点;
    代码:
    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cmath>
    #include <algorithm>
    #include <climits>
    #include <cstring>
    #include <string>
    #include <set>
    #include <map>
    #include <queue>
    #include <stack>
    #include <vector>
    #include <list>
    #define rep(i,m,n) for(i=m;i<=n;i++)
    #define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
    #define mod 1000000007
    #define inf 0x3f3f3f3f
    #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 Lson L, mid, ls[rt]
    #define Rson mid+1, R, rs[rt]
    const int maxn=1e6+10;
    using namespace std;
    ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
    ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
    inline ll read()
    {
        ll x=0;int f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    int n,m,k,t,q,tot,h[maxn],bel[maxn],fa[maxn],son[maxn],sec_son[maxn],dp[maxn],dp1[maxn],ans;
    struct node
    {
        int to,nxt;
    }e[maxn<<1];
    void add(int x,int y)
    {
        tot++;
        e[tot].to=y;
        e[tot].nxt=h[x];
        h[x]=tot;
    }
    void dfs(int now,int pre)
    {
        son[now]=sec_son[now]=inf;
        for(int i=h[now];i;i=e[i].nxt)
        {
            int to=e[i].to;
            if(to!=pre)
            {
                fa[to]=now;
                if(now!=1)bel[to]=bel[now];
                dfs(to,now);
                if(son[now]>min(son[to],to))
                {
                    sec_son[now]=son[now];
                    son[now]=min(son[to],to);
                }
            }
        }
    }
    void dfs1(int now,int pre)
    {
        if(now!=1)dp1[now]=min(dp1[fa[now]],dp[now]);
        else dp1[now]=inf;
        for(int i=h[now];i;i=e[i].nxt)
        {
            int to=e[i].to;
            if(to!=pre)
            {
                dfs1(to,now);
            }
        }
    }
    void solve(int x,int y)
    {
        int now_ans=inf;
        for(int i=h[1];i;i=e[i].nxt)
        {
            int to=e[i].to;
            if(to!=bel[x]&&to!=bel[y])now_ans=min(min(now_ans,son[to]),to);
        }
        if(x!=1)now_ans=min(now_ans,son[x]);
        if(y!=1)now_ans=min(now_ans,son[y]);
        now_ans=min(now_ans,dp1[x]);
        now_ans=min(now_ans,dp1[y]);
        ans=now_ans;
    }
    int main()
    {
        int i,j;
        while(~scanf("%d%d",&n,&q))
        {
            ans=0;
            tot=0;
            memset(h,0,sizeof(h));
            rep(i,1,n)bel[i]=i;
            rep(i,1,n-1)
            {
                int a,b;
                scanf("%d%d",&a,&b);
                add(a,b);
                add(b,a);
            }
            dfs(1,0);
            rep(i,2,n)
            {
                if(fa[i]==1)
                {
                    dp[i]=inf;
                    continue;
                }
                if(min(son[i],i)!=son[fa[i]])dp[i]=son[fa[i]];
                else dp[i]=sec_son[fa[i]];
            }
            dfs1(1,0);
            while(q--)
            {
                int a,b;
                scanf("%d%d",&a,&b);
                a^=ans,b^=ans;
                if(bel[a]==bel[b])ans=1;
                else solve(a,b);
                printf("%d
    ",ans);
            }
        }
        //system("Pause");
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/dyzll/p/5936142.html
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