【BZOJ 1036】【ZJOI 2008】树的统计 树链剖分模板题

sth神犇的模板:

//bzoj1036 题目:一个n个点的树每个点有一个权值,支持修改单点权值,求某两点路径上的点权和或最大点权。
#include <cstdio>
using namespace std;
int pos[30001],f[30001],up[30001],son[30001],size[30001],a[80001],next[80001],last[30001],sum[100001],max[100001];//pos是指某点在线段树中的位置;f是父节点;up是所在链的最上端;son是众儿子中在同一链中的那一个;size是子树的点数;sum是线段树区间和;max是线段树区间最大值
int n,mm,tot,deep[30001],w[30001],q,que,ll,rr,mmax,ssum,which[30001];
char c;//w是单点权值;which是线段树某位置对应的是哪个点,可看做与pos双向映射。
void build(int x,int y)
{
	a[++mm]=y;
	next[mm]=last[x];
	last[x]=mm;
	a[++mm]=x;
	next[mm]=last[y];
	last[y]=mm;
}
void dfs1(int x)
{
	int j,mmax=0,num=0;
	size[x]=1;
	for (j=last[x];j;j=next[j]) if (a[j]!=f[x])
	{
		f[a[j]]=x;
		deep[a[j]]=deep[x]+1;
		dfs1(a[j]);
		size[x]+=size[a[j]];
		if (size[a[j]]>mmax) {mmax=size[a[j]]; num=a[j];}
	}
	son[x]=num;
}
void dfs2(int x)
{
	pos[x]=++tot;
	which[tot]=x;
	if (!son[x]) return;
	up[son[x]]=up[x];
	dfs2(son[x]);
	int j;
	for (j=last[x];j;j=next[j]) if (a[j]!=f[x]&&a[j]!=son[x])
	{
		up[a[j]]=a[j];
		dfs2(a[j]);
	}
}
void buildtree(int now,int l,int r)
{
	if (l==r)
	{
		sum[now]=max[now]=w[which[l]];
		return;
	}
	int x=now<<1;
	buildtree(x,l,(l+r)>>1);
	buildtree(x+1,((l+r)>>1)+1,r);
	if (max[x]>max[x+1]) max[now]=max[x]; else max[now]=max[x+1];
	sum[now]=sum[x]+sum[x+1];
}
void change(int now,int l,int r)
{
	if (l==r)
	{
		max[now]=sum[now]=rr;
		return;
	}
	int mid=(l+r)>>1,x=now<<1;
	if (ll<=mid) change(x,l,mid); else change(x+1,mid+1,r);
	if (max[x]>max[x+1]) max[now]=max[x]; else max[now]=max[x+1];
	sum[now]=sum[x]+sum[x+1];
}
void getans(int now,int l,int r)
{
	if (l>=ll&&r<=rr)
	{
		if (max[now]>mmax) mmax=max[now];
		ssum+=sum[now];
		return;
	}
	int mid=(l+r)>>1;
	if (ll<=mid) getans(now<<1,l,mid);
	if (rr>mid) getans((now<<1)+1,mid+1,r);
}
int main()
{
	freopen("lx.in","r",stdin);
	freopen("lx.out","w",stdout);
	scanf("%d",&n);
	int i,x,y,tmp;
	for (i=1;i<n;i++)
	{
		scanf("%d%d",&x,&y);
		build(x,y);
	}
	deep[1]=1;
	dfs1(1);
	up[1]=1;
	dfs2(1);
	for (i=1;i<=n;i++) scanf("%d",&w[i]);
	buildtree(1,1,n);
	scanf("%d\n",&que);
	for (q=1;q<=que;q++)
	{
		c=getchar();
		if (c=='C')
		{
			for (i=1;i<=5;i++) c=getchar();
			scanf("%d%d\n",&x,&y);//将点x的权值改为r
			ll=pos[x];
			rr=y;
			change(1,1,n);
		}
		else
		{
			for (i=1;i<=3;i++) c=getchar();
			scanf("%d%d\n",&x,&y);
			mmax=-99999999;
			ssum=0;
			while (up[x]!=up[y])
			{
				if (deep[up[x]]<deep[up[y]]) {tmp=x; x=y; y=tmp;}
				ll=pos[up[x]];
				rr=pos[x];
				getans(1,1,n);
				x=f[up[x]];
			}
			if (deep[x]>deep[y]) {tmp=x; x=y; y=tmp;}
			ll=pos[x];
			rr=pos[y];
			getans(1,1,n);
			if (c=='X') printf("%d\n",mmax);else printf("%d\n",ssum);
		}
	}
	return 0;
}

  

 链剖,xyx说是链抛。给xyx神犇跪了O)Z

不说了,(;′⌒`)这是我的链剖模板。为何在codevs上总是RE?我对codevs逐渐产生了隔膜。

#include<cmath>
#include<cctype>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);++i)
#define for4(i,a,n) for(int i=(a);i>(n);++i)
#define CC(i,a) memset(i,a,sizeof(i))
using namespace std;
inline const int max(const int &a,const int &b){return a>b?a:b;}
inline const int min(const int &a,const int &b){return a<b?a:b;}
inline void swapp(int &a,int &b){int c=a;a=b;b=c;}
struct node{int nxt,to;}E[80003];
int N,point[30003],sum[100003],ma[100003],cnt=1,son[30003],up[30003],size[30003],deep[30003];
int pos[30003],which[30003],tot=0,a[30003],fa[30003];
inline void insect(int x,int y){E[cnt].nxt=point[x];E[cnt].to=y;point[x]=cnt++;}
inline void dfs1(int x){
	int nmax=0,num=0;
	size[x]=1;
	for(int tmp=point[x];tmp;tmp=E[tmp].nxt)if (E[tmp].to!=fa[x]){
		fa[E[tmp].to]=x;
		deep[E[tmp].to]=deep[x]+1;
		dfs1(E[tmp].to);
		size[x]+=size[E[tmp].to];
		if (size[E[tmp].to]>nmax){nmax=size[E[tmp].to];num=E[tmp].to;}
	}son[x]=num;
}
inline void dfs2(int x){
	pos[x]=++tot;which[tot]=x;
	if (!son[x]) return;
	up[son[x]]=up[x];
	dfs2(son[x]);
	for(int tmp=point[x];tmp;tmp=E[tmp].nxt)if ((E[tmp].to!=fa[x])&&(E[tmp].to!=son[x])){
		up[E[tmp].to]=E[tmp].to;
		dfs2(E[tmp].to);
	}
}
inline void buildtree(int l,int r,int rt){
	if (l==r){sum[rt]=ma[rt]=a[which[l]];return;}
	int mid=(l+r)>>1;
	buildtree(l,mid,rt<<1);
	buildtree(mid+1,r,rt<<1|1);
	sum[rt]=sum[rt<<1]+sum[rt<<1|1];
	ma[rt]=max(ma[rt<<1],ma[rt<<1|1]);
}
inline void change(int ll,int rr,int l,int r,int rt){
	if (l==r){sum[rt]=ma[rt]=rr;return;}
	int mid=(l+r)>>1;
	if (ll<=mid) change(ll,rr,l,mid,rt<<1);
	else change(ll,rr,mid+1,r,rt<<1|1);
	sum[rt]=sum[rt<<1]+sum[rt<<1|1];
	ma[rt]=max(ma[rt<<1],ma[rt<<1|1]);
}
inline int getmax(int L,int R,int l,int r,int rt){
	if ((L<=l)&&(r<=R)){return ma[rt];}
	int mid=(l+r)>>1,s=-1E9;
	if (L<=mid) s=getmax(L,R,l,mid,rt<<1);
	if (R>mid) s=max(s,getmax(L,R,mid+1,r,rt<<1|1));
	return s;
}
inline void query1(int x,int y){
	int mmax=-1E9;
	while (up[x]!=up[y]){
		if (deep[up[x]]<deep[up[y]])swapp(x,y);
		mmax=max(mmax,getmax(pos[up[x]],pos[x],1,N,1));
		x=fa[up[x]];
	}
	if (deep[x]>deep[y])swapp(x,y);
	mmax=max(mmax,getmax(pos[x],pos[y],1,N,1));
	printf("%d\n",mmax);
}
inline int getsum(int L,int R,int l,int r,int rt){
	if ((L<=l)&&(r<=R)){return sum[rt];}
	int mid=(l+r)>>1,s=0;
	if (L<=mid)s+=getsum(L,R,l,mid,rt<<1);
	if (R>mid)s+=getsum(L,R,mid+1,r,rt<<1|1);
	return s;
}
inline void query2(int x,int y){
	int ssum=0;
	while (up[x]!=up[y]){
		if (deep[up[x]]<deep[up[y]])swapp(x,y);
		ssum+=getsum(pos[up[x]],pos[x],1,N,1);
		x=fa[up[x]];
	}
	if (deep[x]>deep[y])swapp(x,y);
	ssum+=getsum(pos[x],pos[y],1,N,1);
	printf("%d\n",ssum);
}
int main(){
	scanf("%d\n",&N);
	CC(point,0);CC(deep,0);CC(size,0);CC(sum,0);CC(ma,0);CC(up,0);CC(son,0);CC(fa,0);
	for2(i,1,N){
		int x,y;scanf("%d%d\n",&x,&y);
		insect(x,y);insect(y,x);
	}deep[1]=1;
	dfs1(1);
	up[1]=1;
	dfs2(1);
	for1(i,1,N) scanf("%d",&a[i]);
	buildtree(1,N,1);
	int que;scanf("%d\n",&que);
	while (que--){
		char c=getchar();
		if (c=='C'){
			for1(i,1,5)c=getchar();
			int ll,rr;scanf("%d%d\n",&ll,&rr);
			change(pos[ll],rr,1,N,1);
		}else{
			for1(i,1,3)c=getchar();
			int x,y;scanf("%d%d\n",&x,&y);
			if (c=='X') query1(x,y);
			else query2(x,y);
		}
	}return 0;
}

  这样就可以了呢,第一次用CA爷的电脑,键盘真好使。

你可能感兴趣的:(【BZOJ 1036】【ZJOI 2008】树的统计 树链剖分模板题)