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#include #include #define maxn 30005using 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;}