luoguP3302 [SDOI2013]森林

https://www.luogu.org/problemnew/show/P3302

看到查询第 k 小,而且是一颗树,可以联想到在树上的主席树,a 和 b 路径中第 k 小可以通过在 a, b, lca(a, b), fa[lca(a, b)] 四个节点对应的主席树上二分得到

实现主席树是很简单的,连接两个点可以启发式合并,现在最大的问题是连接两个点后还要高效的求出 LCA,一种做法是启发式合并的时候重构倍增数组,也有一种用 LCT 维护 LCA 的方法,相对简单。博主的实现是后者

需要注意的就是找到 fa[lca(a, b)] 的时候,可以模仿 LCT findroot 函数的实现,先 access(lca),然后 splay(lca),找到深度次大的点即可。link 之后 root 会变,进行新操作的时候要 makeroot,才能求出正确的 LCA

总而言之,需要注意的细节还是很多的,具体还是看代码实现吧

#include <bits/stdc++.h>
using namespace std;

typedef unsigned long long ull;
typedef long long ll;

template <typename _T>
inline void read(_T &f) {
    f = 0; _T fu = 1; char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') fu = -1; c = getchar();}
    while(c >= '0' && c <= '9') {f = (f << 3) + (f << 1) + (c & 15); c = getchar();}
    f *= fu;
}

const int N = 80000 + 10;

int fa[N], ch[N][2], rev[N], st[N], len;
int n, m, t;

int isroot(int u) {return ch[fa[u]][0] != u && ch[fa[u]][1] != u;}
int get(int u) {return ch[fa[u]][1] == u;}

void pushdown(int u) {
    if(rev[u]) {
        swap(ch[u][0], ch[u][1]);
        rev[ch[u][0]] ^= 1;
        rev[ch[u][1]] ^= 1;
        rev[u] ^= 1;
    }
}

void rotate(int u) {
    int old = fa[u], oldd = fa[old], k = get(u);
    if(!isroot(old)) ch[oldd][get(old)] = u; fa[u] = oldd;
    fa[ch[u][k ^ 1]] = old; ch[old][k] = ch[u][k ^ 1];
    fa[old] = u; ch[u][k ^ 1] = old;
}

void splay(int u) {
    st[len = 1] = u;
    for(int i = u; !isroot(i); i = fa[i]) st[++len] = fa[i];
    for(int i = len; i >= 1; i--) pushdown(st[i]);
    for(; !isroot(u); rotate(u)) if(!isroot(fa[u])) rotate(get(u) == get(fa[u]) ? fa[u] : u);
}

int access(int u) {
    int tmp;
    for(int i = 0; u; i = u, u = fa[u]) {
        splay(u);
        ch[u][1] = i;
        tmp = u;
    }
    return tmp;
}

void makeroot(int u) {
    access(u);
    splay(u);
    rev[u] ^= 1;
}

void link(int x, int y) {
    makeroot(x);
    fa[x] = y;
}

int LCA(int x, int y) {
    access(x);
    return access(y);
}

int rt[N], val[N * 600], lc[N * 600], rc[N * 600], a[N], b[N];
int tot = 0;

void build(int &u, int l, int r) {
    u = ++tot;
    if(l == r) return;
    int mid = (l + r) >> 1;
    build(lc[u], l, mid);
    build(rc[u], mid + 1, r);
}

void ins(int &u, int pre, int l, int r, int x) {
    u = ++tot;
    val[u] = val[pre] + 1, lc[u] = lc[pre], rc[u] = rc[pre];
    if(l == r) return;
    int mid = (l + r) >> 1;
    if(mid >= x) ins(lc[u], lc[pre], l, mid, x);
    else ins(rc[u], rc[pre], mid + 1, r, x);
}

int query(int a, int b, int c, int d, int l, int r, int k) {
    if(l == r) return l;
    int lsum = val[lc[a]] + val[lc[b]] - val[lc[c]] - val[lc[d]];
    int rsum = val[rc[a]] + val[rc[b]] - val[rc[c]] - val[rc[d]];
    int mid = (l + r) >> 1;
    if(lsum >= k) return query(lc[a], lc[b], lc[c], lc[d], l, mid, k);
    else return query(rc[a], rc[b], rc[c], rc[d], mid + 1, r, k - lsum);
}

int blen;

vector <int> g[N];
int siz[N], head[N];

int f[N]; int find(int x) {return f[x] == x ? x : f[x] = find(f[x]);}

inline void addedge(int u, int v) {
    g[u].push_back(v);
    g[v].push_back(u);
}

void dfs1(int u, int f) {
    ins(rt[u], rt[f], 1, blen, a[u]);
    for(vector <int> :: iterator it = g[u].begin(); it != g[u].end(); it++) {
        int v = *it; if(v == f) continue; dfs1(v, u);
    }
}

void Merge(int a, int b) {
    int x = find(a), y = find(b);
    if(siz[x] < siz[y]) swap(a, b), swap(x, y);
    siz[x] += siz[y]; addedge(a, b); 
    link(a, b); f[y] = x;
    dfs1(b, a);
}

int lastans = 0;

int main() {
    cin >> n;
    cin >> n >> m >> t;
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        b[i] = a[i];
        f[i] = i; siz[i] = 1;
    }
    sort(b + 1, b + n + 1);
    blen = unique(b + 1, b + n + 1) - b - 1;
    for(int i = 1; i <= n; i++) a[i] = lower_bound(b + 1, b + blen + 1, a[i]) - b;
    build(rt[0], 1, blen);
    for(int i = 1; i <= n; i++) {
        ins(rt[i], rt[0], 1, blen, a[i]);
    }
    for(int i = 1; i <= m; i++) {
        int A, B;
        scanf("%d %d", &A, &B);
        Merge(A, B);
    }
    for(int i = 1; i <= t; i++) {
        char c = getchar();
        while(c != 'Q' && c != 'L') c = getchar();
        if(c == 'L') {
            int A, B;
            scanf("%d %d", &A, &B);
            A ^= lastans; B ^= lastans;
            Merge(A, B);
        } else {
            int A, B, K;
            scanf("%d %d %d", &A, &B, &K);
            A ^= lastans; B ^= lastans; K ^= lastans;
            makeroot(find(A));
            int lca = LCA(A, B), father;
            if(find(A) == lca) father = 0;
            else {
                access(lca); splay(lca);
                father = ch[lca][0];
                while(ch[father][1]) {
					father = ch[father][1];
					pushdown(father);
				}
            }
            printf("%d\n", lastans = b[query(rt[A], rt[B], rt[lca], rt[father], 1, blen, K)]);
        }
    }
    return 0;
}
posted @ 2018-09-06 11:25  LJC00118  阅读(488)  评论(1编辑  收藏  举报
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