bzoj2843极地旅行社题解

• 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;
}