bzoj2002: [Hnoi2010]Bounce 弹飞绵羊

lct入门题？只需要Link Cut，不需要换根和维护其他标记

#include<bits/stdc++.h>

using namespace std;

const int N = 200000 + 10;

int ch[N][2], p[N], sz[N], tl[N];

int n, m;

bool isroot(int x) {
    return ch[p[x]][0] != x && ch[p[x]][1] != x;
}
#define l ch[x][0]
#define r ch[x][1]
void update(int x) {
    assert(x != 0);    
    sz[x] = sz[l] + sz[r] + 1;
}
#undef l
#undef r

void rotate(int x) {
    int y = p[x], z = p[y];
    int l = ch[y][1] == x, r = l ^ 1;
    if(!isroot(y)) ch[z][ch[z][1] == y] = x;
    p[ch[x][r]] = y;
    p[y] = x;
    p[x] = z;

    ch[y][l] = ch[x][r];
    ch[x][r] = y;

    update(y), update(x);
}

void splay(int x) {
    while(!isroot(x)) {
        int y = p[x], z = p[y];
        if(!isroot(y)) {
            if((ch[y][1] == x) ^ (ch[z][1] == y)) rotate(x);
            else rotate(y);
        }
        rotate(x);
    }
}

void access(int x) {
    for(int t = 0; x ; t = x, x = p[x]) {
        splay(x), ch[x][1] = t, update(x);
    }
}

int getroot(int x) {
    for(access(x), splay(x); ch[x][0]; x = ch[x][0]);
    return splay(x), x;
}

void Cut(int x) {
    access(x), splay(x);
    p[ch[x][0]] = 0, ch[x][0] = 0;
    update(x);
}

void Link(int x, int y) {
    Cut(x), p[x] = y;
}

int main() {
#ifdef DEBUG
    freopen("in.txt", "r", stdin);
#endif

    int n; scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        sz[i] = 1, scanf("%d", tl + i);
        p[i] = min(tl[i] + i, n + 1);
        tl[i] = p[i];
    }
    sz[n + 1] = 1;

    int m; scanf("%d", &m);
    while(m--) {
        int cmd, a, b;
        scanf("%d%d", &cmd, &a), ++a;
        if(cmd == 1) {
            access(a), splay(a);
            printf("%d\n", sz[a] - 1);
        } else {
            scanf("%d", &b);
            int y = min(a + b, n + 1);
            if(y != tl[a]) Link(a, tl[a] = y);
        }
    }

    return 0;
}

 

分块也是可以的

每个点下一个维护跳出块的点是哪个，以及要用多少步。

sz取$\sqrt{2n}$比较合适。

#include<bits/stdc++.h>

using namespace std;

const int N = 200000 + 10;

int next[N], step[N], belong[N], fa[N];

void modify(int x, int y) {
    if(belong[x] != belong[y]) {
        step[x] = 1, next[x] = y;
    } else {
        step[x] = step[y] + 1;
        next[x] = next[y];
    }
}

int main() {
#ifdef DEBUG
    freopen("in.txt", "r", stdin);
#endif

    int n; scanf("%d", &n);
    int bs = sqrt(n * 2) + 1;
    for(int i = 0; i < n; i++) {
        belong[i] = i / bs;
        scanf("%d", fa + i);
        fa[i] = min(fa[i] + i, n);
    }
    belong[n] = -1;
    for(int i = n-1; i >= 0; i--) modify(i, min(n, fa[i]));

    int m; scanf("%d", &m);
    while(m--) {
        int opt, x;
        scanf("%d%d", &opt, &x);
        if(opt == 1) {
            int ans = 0;
            while(x != n) {
                ans += step[x];
                x = next[x];
            }
            printf("%d\n", ans);
        } else {
            scanf("%d", fa + x);
            fa[x] = min(fa[x] + x, n);
            for(int i = x; i >= 0 && belong[i] == belong[x]; i--) {
                modify(i, fa[i]);
            }
        }
    }

    return 0;
}

 

splay维护括号序列也是可以的。

第一次写splay维护括号序列，用了好多以前没用过的splay写法。

定位指针自底向上的splay知不知道根是谁简直无所谓，随便splay一个就成根了。

#include<bits/stdc++.h>

using namespace std;

const int N = 200000 + 10;

vector<int> G[N];
int dfn[2][N], dfs_clock;

const int maxnode = 400000 + 10;
int ch[maxnode][2], p[maxnode], val[maxnode], sum[maxnode];

void dfs(int u) {
    val[dfn[0][u] = ++dfs_clock] = 1;
    for(unsigned i = 0; i < G[u].size(); i++) {
        dfs(G[u][i]);
    }
    val[dfn[1][u] = ++dfs_clock] = -1;
}

// splay

void update(int x) {
    sum[x] = val[x];
    sum[x] += sum[ch[x][0]];
    sum[x] += sum[ch[x][1]];
}

void rotate(int x) {
    int y = p[x], z = p[y];
    if(z) ch[z][ch[z][1] == y] = x;
    int l = ch[y][1] == x, r = l ^ 1;
    p[x] = z, p[y] = x, p[ch[x][r]] = y;
    ch[y][l] = ch[x][r], ch[x][r] = y;
    update(y), update(x);
}

void splay(int x, int FA = 0) {
    while(p[x] != FA) {
        int y = p[x], z = p[y];
        if(z) {
            if((ch[y][1] == x) ^ (ch[z][1] == y)) rotate(x);
            else rotate(y);
        }
        rotate(x);
    }
}

int tot = 0;
int build(int sz) {
    if(sz <= 0) return 0;
    int mid = sz >> 1;
    int l = build(mid),
        x = ++tot,
        r = build(sz - mid - 1);
    ch[x][0] = l, ch[x][1] = r;
    p[l] = x, p[r] = x, update(x);
    return x;
}

int get_pre(int x) {
    splay(x), x = ch[x][0];
    while(ch[x][1]) x = ch[x][1];
    return splay(x), x;
}

int get_suf(int x) {
    splay(x), x = ch[x][1];
    while(ch[x][0]) x = ch[x][0];
    return splay(x), x;
}

int split(int x) { //将x作为左边的最后一个元素分裂
    splay(x);
    int y = ch[x][1];
    ch[x][1] = 0, p[y] = 0, update(x);
    return y;
}

bool iot(int x, int y) {
    splay(x), splay(y);
    if(p[x]) return 1;
    return 0;
}

void merge(int x, int y) { //将x所在的子树与y所在的子树合并
    splay(x), splay(y);
    while(ch[x][1]) x = ch[x][1];
    splay(x), ch[x][1] = y, p[y] = x, update(x);
}

int main() {
#ifdef DEBUG
    freopen("in.txt", "r", stdin);
#endif

    int n; scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        int fa; scanf("%d", &fa);
        fa = min(fa + i, n + 1);
        G[fa].push_back(i);
    }

    dfs(n + 1);
    
    {
        int x = (n + 1) * 2 + 1;
        ch[x][1] = x + 1, p[ch[x][1]] = x;
        ch[x+1][0] = build((n + 1) << 1);
        p[ch[x+1][0]] = x+1;
        update(x+1), update(x);
    }

    int m; scanf("%d", &m);
    while(m--) {
        int opt, u, fa;
        scanf("%d%d", &opt, &u); ++u;
        if(opt == 1) {
            int x = get_suf(dfn[0][u]);
            splay(x);
            printf("%d\n", sum[ch[x][0]] - 1);
        } else {
            scanf("%d", &fa);
            fa = min(fa + u, n + 1);
            int l = get_pre(dfn[0][u]);
            split(l);
            int r = split(dfn[1][u]);
            merge(l, r);
            split(dfn[0][fa]);
            merge(dfn[0][fa], dfn[1][u]);
            merge(dfn[1][u], dfn[1][fa]);
        }
    }

    return 0;
}