BZOJ 1036: [ZJOI2008]树的统计Count (树链剖分 + 线段树)
2016-09-16 16:54
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解题思路:树链剖分简单题。
/* ***********************************************
┆ ┏┓ ┏┓ ┆
┆┏┛┻━━━┛┻┓ ┆
┆┃ ┃ ┆
┆┃ ━ ┃ ┆
┆┃ ┳┛ ┗┳ ┃ ┆
┆┃ ┃ ┆
┆┃ ┻ ┃ ┆
┆┗━┓ 马 ┏━┛ ┆
┆ ┃ 勒 ┃ ┆
┆ ┃ 戈 ┗━━━┓ ┆
┆ ┃ 壁 ┣┓┆
┆ ┃ 的草泥马 ┏┛┆
┆ ┗┓┓┏━┳┓┏┛ ┆
┆ ┃┫┫ ┃┫┫ ┆
┆ ┗┻┛ ┗┻┛ ┆
************************************************ */
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <bitset>
using namespace std;
#define rep(i,a,b) for (int i=(a),_ed=(b);i<=_ed;i++)
#define per(i,a,b) for (int i=(b),_ed=(a);i>=_ed;i--)
#define pb push_back
#define mp make_pair
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define LL long long
#define ULL unsigned long long
#define MS0(X) memset((X), 0, sizeof((X)))
#define SelfType int
SelfType Gcd(SelfType p,SelfType q){return q==0?p:Gcd(q,p%q);}
SelfType Pow(SelfType p,SelfType q){SelfType ans=1;while(q){if(q&1)ans=ans*p;p=p*p;q>>=1;}return ans;}
#define Sd(X) int (X); scanf("%d", &X)
#define Sdd(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define Sddd(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define reunique(v) v.resize(std::unique(v.begin(), v.end()) - v.begin())
#define all(a) a.begin(), a.end()
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef vector<int> vi;
typedef vector<long long> vll;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")
const int N = 30005;
struct Edge
{
int to,nx;
}edge[N*2];
int head
,cnt;
void addedge(int u,int v)
{
edge[cnt] = Edge{v,head[u]};
head[u] = cnt++;
}
/*树链剖分*/
int top
,siz
,son
;
int fa
,dep
,tot,fp
,tid
;
/*
top[v]表示v所在的链的顶端节点,
son[v]表示与v在同一重链上的v的儿子节点
dep[v]表示v的深度(根深度为1),
siz[v]表示以v为根的子树的节点数,
fa[v]表示v的父亲,
fp[id]=v表示在线段树上编号为id的是节点v
tid[v]表示在线段树上的编号
tot 表示编号
*/
void dfs1(int u,int father,int d)
{
dep[u] = d;
fa[u] = father;
siz[u] = 1;
for(int i=head[u];i;i=edge[i].nx)
{
int v = edge[i].to;
if(v==father)continue;
dfs1(v,u,d+1);
siz[u] += siz[v];
if(siz[v]==-1 || siz[v]>siz[son[u]])
son[u] = v;
}
}
void dfs2(int u,int tp)
{
top[u] = tp;
tid[u] = ++tot;
fp[tot] = u;
if(son[u]==-1)return;
dfs2(son[u],tp);
for(int i=head[u];i;i=edge[i].nx)
{
int v = edge[i].to;
if(v==son[u]|| v== fa[u])continue;
dfs2(v,v);
}
}
/*线段树*/
struct node
{
int l,r,sum,mx;
}seg[N*4];
void pushup(int rt)
{
seg[rt].sum = seg[rt<<1].sum + seg[rt<<1|1].sum;
seg[rt].mx = max(seg[rt<<1].mx,seg[rt<<1|1].mx);
}
void build(int l,int r,int rt)
{
seg[rt].l = l, seg[rt].r = r;
if(l==r) return;
int mid = (l+r) >> 1;
build(l,mid,rt<<1);
build(mid+1,r,rt<<1|1);
}
void update(int x,int y,int rt)
{
int l = seg[rt].l, r = seg[rt].r;
if(l==r)
{
seg[rt].mx = seg[rt].sum = y;
return;
}
int mid = (l+r) >> 1;
if(x<=mid) update(x,y,rt<<1);
else update(x,y,rt<<1|1);
pushup(rt);
}
int query_sum(int L,int R,int rt)
{
int l = seg[rt].l, r = seg[rt].r;
if(L<=l && r<=R)
{
return seg[rt].sum;
}
int mid = (l+r) >> 1;
int ret = 0;
if(L<=mid) ret += query_sum(L,R,rt<<1);
if(R>mid) ret += query_sum(L,R,rt<<1|1);
return ret;
}
int query_max(int L,int R,int rt)
{
int l = seg[rt].l, r = seg[rt].r;
if(L<=l && r<=R)
{
return seg[rt].mx;
}
int mid = (l+r) >> 1;
int ret = -inf_int;
if(L<=mid) ret = max(ret,query_max(L,R,rt<<1));
if(R>mid) ret = max(ret,query_max(L,R,rt<<1|1));
return ret;
}
int solve_sum(int x,int y)
{
int sum = 0;
while(top[x]!=top[y])
{
if(dep[top[x]]<dep[top[y]])swap(x,y);
sum += query_sum(tid[top[x]],tid[x],1);
x = fa[top[x]];
}
if(dep[x]>dep[y])swap(x,y);
sum += query_sum(tid[x],tid[y],1);
return sum;
}
int solve_max(int x,int y)
{
int mx = -inf_int;
while(top[x]!=top[y])
{
if(dep[top[x]]<dep[top[y]])swap(x,y);
mx = max(query_max(tid[top[x]],tid[x],1),mx);
x = fa[top[x]];
}
if(dep[x]>dep[y])swap(x,y);
mx = max(query_max(tid[x],tid[y],1),mx);
return mx;
}
int val
;
void init()
{
MS0(head);
memset(son,-1,sizeof son);
cnt = 1;
tot = 0;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
ios::sync_with_stdio(0);
cin.tie(0);
init();
int n;
n = read();
for(int i=1;i<n;i++)
{
int u,v;
u = read(), v = read();
addedge(u,v);
addedge(v,u);
}
for(int i=1;i<=n;i++)
{
val[i] = read();
}
dfs1(1,0,1);
dfs2(1,1);
build(1,n,1);
for(int i=1;i<=n;i++)
{
update(tid[i],val[i],1);
}
int q;
q = read();
char s[10];
while(q--)
{
int x,y;
scanf("%s%d%d",s,&x,&y);
if(s[0]=='C')
{
val[x] = y;
update(tid[x],y,1);
}
else
{
if(s[1]=='M')
printf("%d\n",solve_max(x,y));
else
printf("%d\n",solve_sum(x,y));
}
}
return 0;
}
解题思路:树链剖分简单题。
/* ***********************************************
┆ ┏┓ ┏┓ ┆
┆┏┛┻━━━┛┻┓ ┆
┆┃ ┃ ┆
┆┃ ━ ┃ ┆
┆┃ ┳┛ ┗┳ ┃ ┆
┆┃ ┃ ┆
┆┃ ┻ ┃ ┆
┆┗━┓ 马 ┏━┛ ┆
┆ ┃ 勒 ┃ ┆
┆ ┃ 戈 ┗━━━┓ ┆
┆ ┃ 壁 ┣┓┆
┆ ┃ 的草泥马 ┏┛┆
┆ ┗┓┓┏━┳┓┏┛ ┆
┆ ┃┫┫ ┃┫┫ ┆
┆ ┗┻┛ ┗┻┛ ┆
************************************************ */
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <bitset>
using namespace std;
#define rep(i,a,b) for (int i=(a),_ed=(b);i<=_ed;i++)
#define per(i,a,b) for (int i=(b),_ed=(a);i>=_ed;i--)
#define pb push_back
#define mp make_pair
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define LL long long
#define ULL unsigned long long
#define MS0(X) memset((X), 0, sizeof((X)))
#define SelfType int
SelfType Gcd(SelfType p,SelfType q){return q==0?p:Gcd(q,p%q);}
SelfType Pow(SelfType p,SelfType q){SelfType ans=1;while(q){if(q&1)ans=ans*p;p=p*p;q>>=1;}return ans;}
#define Sd(X) int (X); scanf("%d", &X)
#define Sdd(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define Sddd(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define reunique(v) v.resize(std::unique(v.begin(), v.end()) - v.begin())
#define all(a) a.begin(), a.end()
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef vector<int> vi;
typedef vector<long long> vll;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")
const int N = 30005;
struct Edge
{
int to,nx;
}edge[N*2];
int head
,cnt;
void addedge(int u,int v)
{
edge[cnt] = Edge{v,head[u]};
head[u] = cnt++;
}
/*树链剖分*/
int top
,siz
,son
;
int fa
,dep
,tot,fp
,tid
;
/*
top[v]表示v所在的链的顶端节点,
son[v]表示与v在同一重链上的v的儿子节点
dep[v]表示v的深度(根深度为1),
siz[v]表示以v为根的子树的节点数,
fa[v]表示v的父亲,
fp[id]=v表示在线段树上编号为id的是节点v
tid[v]表示在线段树上的编号
tot 表示编号
*/
void dfs1(int u,int father,int d)
{
dep[u] = d;
fa[u] = father;
siz[u] = 1;
for(int i=head[u];i;i=edge[i].nx)
{
int v = edge[i].to;
if(v==father)continue;
dfs1(v,u,d+1);
siz[u] += siz[v];
if(siz[v]==-1 || siz[v]>siz[son[u]])
son[u] = v;
}
}
void dfs2(int u,int tp)
{
top[u] = tp;
tid[u] = ++tot;
fp[tot] = u;
if(son[u]==-1)return;
dfs2(son[u],tp);
for(int i=head[u];i;i=edge[i].nx)
{
int v = edge[i].to;
if(v==son[u]|| v== fa[u])continue;
dfs2(v,v);
}
}
/*线段树*/
struct node
{
int l,r,sum,mx;
}seg[N*4];
void pushup(int rt)
{
seg[rt].sum = seg[rt<<1].sum + seg[rt<<1|1].sum;
seg[rt].mx = max(seg[rt<<1].mx,seg[rt<<1|1].mx);
}
void build(int l,int r,int rt)
{
seg[rt].l = l, seg[rt].r = r;
if(l==r) return;
int mid = (l+r) >> 1;
build(l,mid,rt<<1);
build(mid+1,r,rt<<1|1);
}
void update(int x,int y,int rt)
{
int l = seg[rt].l, r = seg[rt].r;
if(l==r)
{
seg[rt].mx = seg[rt].sum = y;
return;
}
int mid = (l+r) >> 1;
if(x<=mid) update(x,y,rt<<1);
else update(x,y,rt<<1|1);
pushup(rt);
}
int query_sum(int L,int R,int rt)
{
int l = seg[rt].l, r = seg[rt].r;
if(L<=l && r<=R)
{
return seg[rt].sum;
}
int mid = (l+r) >> 1;
int ret = 0;
if(L<=mid) ret += query_sum(L,R,rt<<1);
if(R>mid) ret += query_sum(L,R,rt<<1|1);
return ret;
}
int query_max(int L,int R,int rt)
{
int l = seg[rt].l, r = seg[rt].r;
if(L<=l && r<=R)
{
return seg[rt].mx;
}
int mid = (l+r) >> 1;
int ret = -inf_int;
if(L<=mid) ret = max(ret,query_max(L,R,rt<<1));
if(R>mid) ret = max(ret,query_max(L,R,rt<<1|1));
return ret;
}
int solve_sum(int x,int y)
{
int sum = 0;
while(top[x]!=top[y])
{
if(dep[top[x]]<dep[top[y]])swap(x,y);
sum += query_sum(tid[top[x]],tid[x],1);
x = fa[top[x]];
}
if(dep[x]>dep[y])swap(x,y);
sum += query_sum(tid[x],tid[y],1);
return sum;
}
int solve_max(int x,int y)
{
int mx = -inf_int;
while(top[x]!=top[y])
{
if(dep[top[x]]<dep[top[y]])swap(x,y);
mx = max(query_max(tid[top[x]],tid[x],1),mx);
x = fa[top[x]];
}
if(dep[x]>dep[y])swap(x,y);
mx = max(query_max(tid[x],tid[y],1),mx);
return mx;
}
int val
;
void init()
{
MS0(head);
memset(son,-1,sizeof son);
cnt = 1;
tot = 0;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
ios::sync_with_stdio(0);
cin.tie(0);
init();
int n;
n = read();
for(int i=1;i<n;i++)
{
int u,v;
u = read(), v = read();
addedge(u,v);
addedge(v,u);
}
for(int i=1;i<=n;i++)
{
val[i] = read();
}
dfs1(1,0,1);
dfs2(1,1);
build(1,n,1);
for(int i=1;i<=n;i++)
{
update(tid[i],val[i],1);
}
int q;
q = read();
char s[10];
while(q--)
{
int x,y;
scanf("%s%d%d",s,&x,&y);
if(s[0]=='C')
{
val[x] = y;
update(tid[x],y,1);
}
else
{
if(s[1]=='M')
printf("%d\n",solve_max(x,y));
else
printf("%d\n",solve_sum(x,y));
}
}
return 0;
}
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