poj 2763 Housewife Wind 【树链剖分维护树上权值和】

题目：poj 2763 Housewife Wind

题意：给一个数，边之间有权值，然后两种操作，第一种：求任意两点的权值和，第二，修改树上两点的权值。

分析：很基础的一个树链剖分维护树上权值和，第二道树链剖分题目，也错了好几次。而且这个题目卡vector。有点坑。

AC代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
#define Del(a,b) memset(a,b,sizeof(a))
const int N = 100005;

int dep[N],siz[N],fa[N],id[N],son[N],val[N],top[N]; //top 最近的重链父节点
int num,head[N],cnt;
struct Edge
{
    int to,next;
};
Edge v[N*2];
struct tree
{
    int x,y,val;
    void read(){
        scanf("%d%d%d",&x,&y,&val);
    }
};
void add_Node(int x,int y)
{
    v[cnt].to = y;
    v[cnt].next = head[x];
    head[x] = cnt++;
}
tree e[N];
void dfs1(int u, int f, int d) {
    dep[u] = d;
    siz[u] = 1;
    son[u] = 0;
    fa[u] = f;
    for(int i = head[u]; i != -1 ;i = v[i].next)
    {
        int to = v[i].to;
        if(to != f)
        {
            dfs1(to,u,d+1);
            siz[u] += siz[to];
            if(siz[son[u]] < siz[to])
                son[u] = to;
        }
    }
}
void dfs2(int u, int tp) {
    top[u] = tp;
    id[u] = ++num;
    if (son[u]) dfs2(son[u], tp);
    for(int i = head[u]; i != -1 ;i = v[i].next )
    {
        int to = v[i].to;
        if(to == fa[u] || to == son[u])
            continue;
        dfs2(to,to);
    }
}
#define lson(x) ((x<<1))
#define rson(x) ((x<<1)+1)
struct Tree
{
    int l,r,val;
};
Tree tree[4*N];
void pushup(int x) {
    tree[x].val = tree[lson(x)].val + tree[rson(x)].val;
}

void build(int l,int r,int v)
{
    tree[v].l=l;
    tree[v].r=r;
    if(l==r){
        tree[v].val = val[l];
        return ;
    }
    int mid=(l+r)>>1;
    build(l,mid,v*2);
    build(mid+1,r,v*2+1);
    pushup(v);
}
void update(int o,int v,int val)  //log(n)
{
    if(tree[o].l==tree[o].r)
    {
        tree[o].val = val;
        return ;
    }
    int mid = (tree[o].l+tree[o].r)/2;
    if(v<=mid)
        update(o*2,v,val);
    else
        update(o*2+1,v,val);
    pushup(o);
}
int query(int o,int l, int r)
{
    if (tree[o].l >= l && tree[o].r <= r) {
        return tree[o].val;
    }
    int mid = (tree[o].l + tree[o].r) / 2;
    if(r<=mid)
        return query(o+o,l,r);
    else if(l>mid)
        return query(o+o+1,l,r);
    else
        return query(o+o,l,mid) + query(o+o+1,mid+1,r);
}

int Yougth(int u, int v) {
    int tp1 = top[u], tp2 = top[v];
    int ans = 0;
    while (tp1 != tp2) {
        if (dep[tp1] < dep[tp2]) {
            swap(tp1, tp2);
            swap(u, v);
        }
        ans += query(1,id[tp1], id[u]);
        u = fa[tp1];
        tp1 = top[u];
    }
    if (u == v) return ans;
    if (dep[u] > dep[v]) swap(u, v);
    ans += query(1,id[son[u]], id[v]);
    return ans;
}
int main()
{
    int n,m,s;
    while(~scanf("%d%d%d",&n,&m,&s))
    {
        cnt = 0;
        memset(head,-1,sizeof(head));
        for(int i=1;i<n;i++)
        {
            e[i].read();
            add_Node(e[i].x,e[i].y);
            add_Node(e[i].y,e[i].x);
        }
        num = 0;
        dfs1(1,0,1);
        dfs2(1,1);
        for(int i=1;i<n;i++)
        {
            if(dep[e[i].x] < dep[e[i].y])
                swap(e[i].x,e[i].y);
            val[id[e[i].x]] = e[i].val;
        }
        build(1,num,1);
        for(int i=0;i<m;i++)
        {
            int ok,x,y;
            scanf("%d",&ok);
            if(ok==0)
            {
                scanf("%d",&x);
                printf("%d\n",Yougth(s,x));
                s = x;
            }
            else
            {
                scanf("%d%d",&x,&y);
                update(1,id[e[x].x],y);
            }
        }
    }
    return 0;
}