SPOJ QTREE 树链剖分

 

　　树链剖分的第一题，易懂，注意这里是边。

#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 99999999
#define ll __int64
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int MAXN = 10010;
struct node
{
    int to;
    int v;
    int next;
}edge[MAXN*3];
int pre[MAXN],ind,top[MAXN],fa[MAXN],son[MAXN],w[MAXN],deq[MAXN],siz[MAXN],fn;
int tree[MAXN<<2],val[MAXN][3],n;
void add(int x,int y,int z)
{
    edge[ind].to = y;
    edge[ind].v = z;
    edge[ind].next = pre[x];
    pre[x] = ind++;
}

//第一次搜索找出siz[],son[],deq[],fa[]
void dfs1(int rt,int pa,int d)
{
    deq[rt] = d;
    son[rt] = 0;
    fa[rt] = pa;
    siz[rt] = 1;
    int i;
    for(i=pre[rt]; i!=-1; i=edge[i].next){
        int t = edge[i].to;
        if(t != fa[rt]){
            dfs1(t,rt,d+1);
            siz[rt] += siz[t];
            if(siz[son[rt]] < siz[t]){//如果父节点的son[rt]的iz[]小于子节点的siz[]更新son[]
                son[rt] = t;
            }
        }
    }
}

//第二次搜索找到w[],top[]
void dfs2(int rt,int tp)
{
    top[rt] = tp;
    w[rt] = ++fn;

    if(son[rt] != 0)
        dfs2(son[rt],tp);//如果当前父节点纯在son 那么为了让链在线段树中连续 先对son[]进行搜索，并且重边顶端的点相同。

    for(int i=pre[rt]; i!=-1; i=edge[i].next){
        int t = edge[i].to;
        if(t != fa[rt] && son[rt] != t){
            dfs2(t,t);
        }
    }
}
void updata(int p,int v,int l,int r,int rt)
{
    if(l == r){
        tree[rt] = v;
        return ;
    }
    int m = (l+r)/2;
    if(m >= p){
        updata(p,v,lson);
    }
    else {
        updata(p,v,rson);
    }
    tree[rt] = max(tree[rt<<1],tree[rt<<1|1]);
}
int query(int L,int R,int l,int r,int rt)
{
    if(L<=l && r<=R){
        return tree[rt];
    }
    int m = (l+r)/2;
    int ans = 0;
    if(m >= L){
        ans = max(ans,query(L,R,lson));
    }
    if(m < R){
        ans = max(ans,query(L,R,rson));
    }
    return ans;
}
int lca(int x,int y)
{
    int f1,f2;
    int ans = 0;
    while(top[x] != top[y])
    {
        if(deq[top[x]] < deq[top[y]]){
            swap(x,y);
        }
        ans = max(ans,query(w[top[x]],w[x],1,fn,1));

        x = fa[top[x]];
    }

    if(x == y)
        return ans;

    if(deq[x] < deq[y]){
        swap(x,y);
    }
    return max(ans,query(w[son[y]],w[x],1,fn,1))    ;
}
int main()
{
    int i,j,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        ind = 1;
        memset(pre,-1,sizeof(pre));
        for(i=1; i<n; i++){
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            val[i][0] = x;
            val[i][1] = y;
            val[i][2] = z;
            add(x,y,z);
            add(y,x,z);
        }
        memset(tree,0,sizeof(tree));
        dfs1(1,1,1);
        
        fn = 0;
        dfs2(1,1);

        for(i=1; i<n; i++){
            if(deq[val[i][0]] > deq[val[i][1]]){
                swap(val[i][0],val[i][1]);
            }
            updata(w[val[i][1]],val[i][2],1,fn,1);
        }
        char s[15];
        while(1)
        {
            scanf("%s",s);
            if(s[0] == 'D')
                break;

            if(s[0] == 'C'){
                int x,y;
                scanf("%d%d",&x,&y);
                updata(w[val[x][1]],y,1,fn,1);
            }
            else {
                int x,y;
                scanf("%d%d",&x,&y);
                printf("%d\n",lca(x,y));
            }
        }
    }
}