• 题目大意
  有n座小岛，当中每一个岛都有若干帝企鹅。

  一開始岛与岛之间互不相连。有m个操作。各自是在两个岛之间修一座双向桥，若两岛已连通则不修并输出no，若不连通就输出yes并修建。改动一个岛上帝企鹅的数量；询问从岛A到岛B可看到多少帝企鹅，若到不了输出impossible。

• 题解
  继续试水LCT。LCT维护每一个点自身的企鹅数以及其在Splay下的子树的企鹅数的总和。

  修桥操作要在LCT中询问是否有同样的根，没有则添边。改动时把要被改动的点弄到树根去，直接改动就可以。查询时要先推断是否为同一结点，再推断是否连通。若连通，则把当中一个点x弄到根上，还有一个点y用access连上去，再用Splay把y弄到辅助树的根上。

  这时x一定是y最左端的子孙（由于它是根。中序序列里最靠前）。直接输出y左子树企鹅的总和加上y自己的企鹅数就可以。

• Code

#include <cstdio>
#include <algorithm>
#include <cstring>
#define maxn 30005
using namespace std;
int n, m;
struct node *nil, *T[maxn], *S[maxn];
struct node
{bool rev;int val, s;node *fa, *lc, *rc;node(bool rev = false, int val = 0, int s = 0, node *fa = nil, node *lc = nil, node *rc = nil): rev(rev), val(val), s(s), fa(fa), lc(lc), rc(rc) {}inline void update(){s = lc -> s + rc -> s + val;}inline void rever(){rev ^= 1;swap(lc, rc);}inline void pushdown(){if(rev){rev = false;lc -> rever(); rc -> rever();}}
};
inline void zig(node *x)
{node *y = x -> fa;y -> lc = x -> rc;x -> rc -> fa = y;x -> rc = y;x -> fa = y -> fa;if(y == y -> fa -> lc) y -> fa -> lc = x;else if(y == y -> fa -> rc) y -> fa -> rc = x;y -> fa = x;y -> update();
}
inline void zag(node *x)
{node *y = x -> fa;y -> rc = x -> lc;x -> lc -> fa = y;x -> lc = y;x -> fa = y -> fa;if(y == y -> fa -> lc) y -> fa -> lc = x;else if(y == y -> fa -> rc) y -> fa -> rc = x;y -> fa = x;y -> update();
}
void splay(node *x)
{int top = 0;S[top++] = x;for(node *i = x; i == i -> fa -> lc || i == i -> fa -> rc; i = i -> fa){S[top++] = i -> fa;}while(top--) S[top] -> pushdown();node *y = nil, *z = nil;while(x == x -> fa -> lc || x == x -> fa -> rc){y = x -> fa; z = y -> fa;if(x == y -> lc){if(y == z -> lc) zig(y);zig(x);}else{if(y == z -> rc) zag(y);zag(x);}}x -> update();
}
inline void access(node *x)
{for(node *y = nil; x != nil; y = x, x = x -> fa){splay(x);x -> rc = y;x -> update();}
}
inline void makeroot(node *x)
{access(x); splay(x); x -> rever();
}
inline void lnk(node *x, node *y)
{makeroot(x);x -> fa = y;splay(y);
}
inline node* find(node *x)
{access(x); splay(x);while(x -> lc != nil) x = x -> lc;return x;
}
inline void change(node *x, int k)
{makeroot(x);x -> val = k;x -> update();
}
inline int query(node *x, node *y)
{if(x == y) return x -> val;if(find(x) != find(y)) return -1;makeroot(x);access(y);splay(y);return (y -> lc -> s + y -> val);
}
int main()
{int a, b, c;char ch[10];scanf("%d", &n);nil = new node(); *nil = node();for(int i = 1; i <= n; ++i) T[i] = new node();for(int i = 1; i <= n; ++i){scanf("%d", &(T[i] -> val));T[i] -> update();}scanf("%d", &m);while(m--){scanf("%s%d%d", ch, &a, &b);if(ch[0] == 'b'){if(find(T[a]) != find(T[b])){lnk(T[a], T[b]); puts("yes");}else puts("no");}else if(ch[0] == 'p'){change(T[a], b);}else{c = query(T[a], T[b]);if(c == -1) puts("impossible");else printf("%d\n", c);}}for(int i = 1; i <= n; ++i){delete T[i];T[i] = NULL;}delete nil;nil = NULL;return 0;
}