bzoj1180: [CROATIAN2009]OTOCI

题目大意：询问两点是否连通（反人类的是连通输no，不联通输yes...），单点权值修改，路径和。

思路：正常的动态树，搞搞就行了。

#include<cstdio>
#include<cstring>
#include<algorithm>
const int maxn=30010;
using namespace std;
int n,m;char op[15];
struct LCT{
	int sum[maxn],fa[maxn],c[maxn][2],rev[maxn],val[maxn];
	bool isroot(int x){return (c[fa[x]][0]!=x)&&(c[fa[x]][1]!=x);}
	int which(int x){return c[fa[x]][1]==x;}
	void update(int x){sum[x]=sum[c[x][0]]+sum[c[x][1]]+val[x];}
	void flip(int x){swap(c[x][0],c[x][1]),rev[x]^=1;}
	void down(int x){if (rev[x]) flip(c[x][0]),flip(c[x][1]),rev[x]=0;}
	void relax(int x){if (!isroot(x)) relax(fa[x]);down(x);}
	void rotate(int x){
		int y=fa[x],z=fa[y],nx=which(x),ny=which(y);
		fa[c[x][!nx]]=y,c[y][nx]=c[x][!nx];
		fa[x]=z;if (!isroot(y)) c[z][ny]=x;
		fa[y]=x,c[x][!nx]=y;update(y);
	}
	void splay(int x){
		relax(x);
		while (!isroot(x)){
			if (isroot(fa[x])) rotate(x);
			else if (which(x)==which(fa[x])) rotate(fa[x]),rotate(x);
			else rotate(x),rotate(x);
		}
		update(x);
	}
	void access(int x){for (int p=0;x;x=fa[x]) splay(x),fa[c[x][1]=p]=x,update(x),p=x;}
	void makeroot(int x){access(x),splay(x),flip(x);}
	void modify(int x,int v){val[x]=v,splay(x);}
	int findroot(int x){access(x),splay(x);for (;c[x][0];x=c[x][0]);return x;}
	void link(int a,int b){makeroot(a),fa[a]=b;}
	void qsum(int a,int b){
		if (findroot(a)!=findroot(b)) return puts("impossible"),void();
		makeroot(a),access(b),splay(b),printf("%d\n",sum[b]);
	}
	void bridge(int a,int b){
		if (findroot(a)==findroot(b)) return puts("no"),void();
		else link(a,b),puts("yes");
	}
}T;

int main(){
	scanf("%d",&n);
	for (int i=1,a;i<=n;i++) scanf("%d",&a),T.modify(i,a);
	scanf("%d",&m);
	for (int i=1,a,b;i<=m;i++){
		scanf("%s%d%d",op,&a,&b);
		if (op[0]=='b') T.bridge(a,b);
		if (op[0]=='p') T.modify(a,b);
		if (op[0]=='e') T.qsum(a,b);
	}
	return 0;
}