【BZOJ2843】极地旅行社

来源：互联网发布：做名片的软件编辑：程序博客网时间：2024/04/27 21:16

题意：连边，询问权值和，单点修改权值

裸的不能再裸的lct模板题。。。话说我就贴了个刚写完的板，改了下输入，输出，数据范围都没看，居然就a了。。。。一点坑点都没有，要不要这样啊。。。

#include<iostream>#include<cstdio>#include<algorithm>#define N 200005using namespace std;int n,m;int read(){    int x=0,f=1;char c=getchar();    for (;c<'0'||c>'9';c=getchar())if(c=='-') f=-1;    for (;c>='0'&&c<='9';c=getchar())x=x*10+c-'0';    return x*f;}struct LCT{int son[N][2],father[N];int val[N],sum[N],Max[N];bool rev[N];int find(int x){access(x);splay(x);while (son[x][0]) x=son[x][0];return x;}bool isroot(int x){return son[father[x]][0]!=x&&son[father[x]][1]!=x;}void up(int x){int l=son[x][0],r=son[x][1];sum[x]=val[x]+sum[l]+sum[r];Max[x]=max(val[x],max(Max[l],Max[r]));}void down(int x){if (!x||!rev[x]) return;rev[son[x][0]]^=1,rev[son[x][1]]^=1;swap(son[x][0],son[x][1]);rev[x]=0;}void pushdown(int x){if (!isroot(x)) pushdown(father[x]);down(x);}void rotate(int x){int y=father[x],z=father[y],l=(son[y][0]!=x),r=l^1;if (!isroot(y)) son[z][son[z][1]==y]=x;father[x]=z;father[y]=x;father[son[x][r]]=y;son[y][l]=son[x][r];son[x][r]=y;up(y);up(x);}void splay(int x){for (pushdown(x);!isroot(x);rotate(x)){int y=father[x],z=father[y];if (!isroot(y))rotate((x==son[y][0])==(y==son[z][0])?y:x);}}void access(int x){for (int t=0;x;t=x,x=father[x])splay(x),son[x][1]=t,up(x);}void makeroot(int x){access(x);splay(x);rev[x]^=1;}void link(int x,int y){makeroot(x);father[x]=y;}void cut(int x,int y){makeroot(x);access(y);splay(y);son[y][0]=father[x]=0;up(y);}void mdy(int x,int y){splay(x);val[x]=y;up(x);}int qry(int x,int y){makeroot(x);access(y);splay(y);return sum[y];}}T;int main(){n=read();for (int i=1;i<=n;i++)T.val[i]=read(),T.sum[i]=T.Max[i]=T.val[i];m=read();while (m--){char p[10];int x,y;scanf("%s%d%d",p,&x,&y);switch(p[0]){case 'e':if (T.find(x)==T.find(y))printf("%d\n",T.qry(x,y));elseputs("impossible");break;case 'b':if (T.find(x)!=T.find(y))T.link(x,y),puts("yes");elseputs("no");break;case 'p':T.mdy(x,y);break;}}return 0;}

0 0