BZOJ 1036 树链剖分模板题

Posted 2020-08-31 掉血菜鸡煮熟中

BZOJ 1036

题意：一棵树，每个点有权值，三种操作：修改一个点的值；询问一条链上最大值；询问一条链上权值和。

tags：模板题

// bzoj 1036
#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define FF(i,a,b) for (int i=a;i<=b;i++)
#define F(i,b,a)  for (int i=b;i>=a;i--)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
typedef long long ll;
const int N = 3e4+10;

struct Edge{int to, next;}e[N];
struct Point{int mx, sum;}t[N<<2];
int n, sz, Size[N], dep[N], w[N], head[N], pos[N], tot, bl[N], fa[N];
void Addedge(int u, int v) { e[tot].to=v, e[tot].next=head[u], head[u]=tot++; }
void Init()
{
    mes(head,-1);
    scanf("%d", &n);
    int u, v;
    FF(i,1,n-1) {
        scanf("%d %d", &u, &v);
        Addedge(u, v);  Addedge(v, u);
    }
}
void dfs1(int u)   //搜索出各个子树的Size
{
    Size[u]=1;
    for(int i=head[u]; i!=-1; i=e[i].next) if(e[i].to!=fa[u]) {
        int v=e[i].to;
        fa[v]=u, dep[v]=dep[u]+1;
        dfs1(v);
        Size[u]+=Size[v];
    }
}
void dfs2(int u, int chain)    //构建树链
{
    sz++, pos[u]=sz, bl[u]=chain;   //pos[]分配u结点在线段树中的编号，chain是一条链上头结点的标号
    int mx=0;
    for(int i=head[u]; i!=-1; i=e[i].next) {
        int v=e[i].to;
        if(dep[v]>dep[u] && Size[v]>Size[mx]) mx=v;
    }
    if(mx==0) return ;
    dfs2(mx, chain);
    for(int i=head[u]; i!=-1; i=e[i].next) {
        int v=e[i].to;
        if(dep[v]>dep[u] && mx!=v) dfs2(v, v);
    }
}
void update(int ro, int l, int r, int x, int y)    //线段树单点更新
{
    if(l==r) { t[ro].mx=t[ro].sum=y; return ; }
    int mid=(l+r)>>1;
    if(x<=mid) update(ro<<1, l, mid, x, y);
    else update(ro<<1|1, mid+1, r, x, y);
    t[ro].mx=max(t[ro<<1].mx, t[ro<<1|1].mx);
    t[ro].sum=t[ro<<1].sum+t[ro<<1|1].sum;
}
int querymx(int ro, int l, int r, int x, int y)     //线段树区间求最大值
{

    if(l==x && r==y) return t[ro].mx;
    int mid=(l+r)>>1;
    if(y<=mid) return querymx(ro<<1, l, mid, x, y);
    else if(mid<x) return querymx(ro<<1|1, mid+1, r, x, y);
    else return max(querymx(ro<<1, l, mid, x, mid), querymx(ro<<1|1, mid+1, r, mid+1, y));
}
int solvemx(int x, int y)
{
    int mx=-INF;
    while(bl[x]!=bl[y]) {
        if(dep[bl[x]]<dep[bl[y]]) swap(x, y);
        mx=max(mx, querymx(1, 1, sz, pos[bl[x]], pos[x]));
        x=fa[bl[x]];
    }
    if(pos[x]>pos[y]) swap(x, y);
    mx=max(mx, querymx(1, 1, sz, pos[x], pos[y]));
    return mx;
}
int querysum(int ro, int l, int r, int x, int y)       //线段树区间求和
{
    if(l==x && r==y) return t[ro].sum;
    int mid=(l+r)>>1;
    if(y<=mid) return querysum(ro<<1, l, mid, x, y);
    else if(mid<x) return querysum(ro<<1|1, mid+1, r, x, y);
    else return querysum(ro<<1, l, mid, x, mid)+querysum(ro<<1|1, mid+1, r, mid+1, y);
}
int solvesum(int x, int y)
{
    int sum=0;
    while(bl[x]!=bl[y]) {        //不在一条重链上时，就一边修改，一边往同一条重链上靠
        if(dep[bl[x]]<dep[bl[y]]) swap(x, y);
        sum+=querysum(1, 1, sz, pos[bl[x]], pos[x]);
        x=fa[bl[x]];
    }
    if(pos[x]>pos[y]) swap(x, y);
    sum+=querysum(1, 1, sz, pos[x], pos[y]);
    return sum;
}
void solve()
{
    FF(i,1,n) scanf("%d", &w[i]), update(1, 1, sz, pos[i], w[i]);
    int q, x, y;
    char str[10];
    scanf("%d", &q);
    FF(i,1,q) {
        scanf("%s %d %d", str, &x, &y);
        if(str[0]==‘C‘) w[x]=y, update(1, 1, sz, pos[x], y);
        else if(str[1]==‘M‘) printf("%d\n", solvemx(x, y));
        else printf("%d\n", solvesum(x, y));
    }
}
int main()
{
    Init();
    dfs1(1);
    dfs2(1, 1);
    solve();

    return 0;
}