HDU 3966 Aragorn's Story --树链剖分

题意： 树上路径之间的点统一加减k，查询某点的值

解法：不会LCA的解法，于是用树链剖分了，比较简单的剖分，然后用线段树维护就行了，相当于区间更新，单点查询，查询点 i 的值时，只需在线段树中查询pos[u]位置的值即可。

加IO优化900+ms

代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <cmath>

#include <algorithm>

using namespace std;

#define N 50007



int siz[N];  //子树大小

int son[N];  //重儿子

int dep[N];  //深度

int pos[N];  //在线段树中的位置

int Top[N];  //所在重链的祖先

int fa[N];   //父节点

int head[2*N],tot,POS,n,m,q;

struct Edge

{

    int v,next;

}G[2*N];

int tree[4*N],mark[4*N],a[N];



void init()

{

    POS = tot = 0;

    memset(head,-1,sizeof(head));

    memset(son,-1,sizeof(son));

    memset(tree,0,sizeof(tree));

    memset(mark,0,sizeof(mark));

}



void addedge(int u,int v)

{

    G[tot].v = v, G[tot].next = head[u], head[u] = tot++;

    G[tot].v = u, G[tot].next = head[v], head[v] = tot++;

}



void pushdown(int l,int r,int rt)

{

    int mid = (l+r)/2;

    if(mark[rt])

    {

        mark[2*rt] += mark[rt];

        mark[2*rt+1] += mark[rt];

        tree[2*rt] += mark[rt]*(mid-l+1);

        tree[2*rt+1] += mark[rt]*(r-mid);

        mark[rt] = 0;

    }

}



void update(int l,int r,int aa,int bb,int k,int rt)

{

    if(aa <= l && bb >= r)

    {

        tree[rt] += k*(r-l+1);

        mark[rt] += k;

        return;

    }

    int mid = (l+r)/2;

    if(aa <= mid)

        update(l,mid,aa,bb,k,2*rt);

    if(bb > mid)

        update(mid+1,r,aa,bb,k,2*rt+1);

}



int query(int l,int r,int pos,int rt)

{

    if(l == r)

        return tree[rt];

    pushdown(l,r,rt);

    int mid = (l+r)/2;

    if(pos <= mid) return query(l,mid,pos,2*rt);

    else return query(mid+1,r,pos,2*rt+1);

}



void dfs(int u,int f)

{

    dep[u] = dep[f]+1;

    siz[u] = 1;

    for(int i=head[u];i!=-1;i=G[i].next)

    {

        int v = G[i].v;

        if(v == f) continue;

        fa[v] = u;

        dfs(v,u);

        if(son[u] == -1 || siz[v] > siz[son[u]]) son[u] = v;

        siz[u] += siz[v];

    }

}



void dfs2(int u,int father)

{

    pos[u] = ++POS;

    Top[u] = father;

    if(son[u] != -1) dfs2(son[u],father);

    for(int i=head[u];i!=-1;i=G[i].next)

    {

        int v = G[i].v;

        if(v != fa[u] && v != son[u])

            dfs2(v,v);

    }

}



void Change(int u,int v,int k)

{

    int fx = Top[u], fy = Top[v];

    while(fx != fy)

    {

        if(dep[fx] < dep[fy])

        {

            swap(u,v);

            swap(fx,fy);

        }

        update(1,POS,pos[fx],pos[u],k,1);

        u = fa[fx];

        fx = Top[u];

    }

    if(dep[u] > dep[v]) swap(u,v);

    update(1,POS,pos[u],pos[v],k,1);

}



inline int in()

{

    char ch;

    int a = 0;

    while((ch = getchar()) == ' ' || ch == '\n');

    a += ch - '0';

    while((ch = getchar()) != ' ' && ch != '\n')

    {

        a *= 10;

        a += ch - '0';

    }

    return a;

}



int main()

{

    int u,v,k,i;

    char ss[5];

    while(scanf("%d%d%d",&n,&m,&q)!=EOF)

    {

        init();

        for(i=1;i<=n;i++)

            a[i] = in();

        for(i=1;i<n;i++)

        {

            u = in(), v = in();

            addedge(u,v);

        }

        dep[0] = 0;

        dfs(1,0);

        dfs2(1,1);

        for(i=1;i<=n;i++)

            update(1,POS,pos[i],pos[i],a[i],1);

        while(q--)

        {

            scanf("%s",ss);

            if(ss[0] == 'Q')

            {

                u = in();

                printf("%d\n",query(1,POS,pos[u],1));

            }

            else

            {

                u = in(), v = in(), k = in();

                if(ss[0] == 'I')

                    Change(u,v,k);

                else

                    Change(u,v,-k);

            }

        }

    }

    return 0;

}

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