「洛谷4197」「BZOJ3545」peak【线段树合并】

题目链接

【洛谷】
【BZOJ】没有权限号嘤嘤嘤。题号:3545

题解

窝不会克鲁斯卡尔重构树怎么办???
可以离线乱搞。
我们将所有的操作全都存下来。
为了解决小于等于\(x\)的操作,那么我们按照长度来排一个序。
如果询问和加边长度相同,这加边优先。
对于每一个连通块进行权值线段树。
权值线段树解决\(k\)大的问题。
每一次合并,并查集判联通,线段树暴力合并。
时间复杂度\(O(nlogn)\)

代码

#include <bits/stdc++.h>
using namespace std;
namespace IOstream {
    #define gc getchar
    template <typename T>
    inline void read(T &x) {
        x = 0; T fl = 1; char c = 0;
        for (; c < '0' || c > '9'; c = gc())
            if (c == '-') fl = -1;
        for (; c >= '0' && c <= '9'; c = gc())
            x = (x << 1) + (x << 3) + (c ^ 48);
        x *= fl;
    }
    #undef gc
} using namespace IOstream;
int n, m, q;
const int N = 100000 + 5; 
int val[N], id[N];
namespace seg {
    #define ls(x) tr[x].lc
    #define rs(x) tr[x].rc
    struct node {
        int lc, rc, s; node() { lc = rc = s = 0; }
    } tr[N * 50];
    int tot = 0; 
    void upd(int &k, int l, int r, int val) {
        if (!k) k = ++ tot; 
        tr[k].s = 1; 
        if (l == r) return; 
        int mid = (l + r) >> 1;
        if (val <= mid) upd(ls(k), l, mid, val);
        else upd(rs(k), mid + 1, r, val);
    }
    int kth(int k, int l, int r, int rk) {
        if (l == r) return l;
        int mid = (l + r) >> 1;
        if (rk <= tr[ls(k)].s) return kth(ls(k), l, mid, rk);
        else return kth(rs(k), mid + 1, r, rk - tr[ls(k)].s);
    }
    int merge(int x, int y) {
        if (!x || !y) return x + y;
        if (!ls(x) && !rs(x)) { tr[x].s += tr[y].s; return x; }
        ls(x) = merge(ls(x), ls(y));
        rs(x) = merge(rs(x), rs(y)); 
        tr[x].s = tr[ls(x)].s + tr[rs(x)].s; 
        return x; 
    } 
}
struct ASK {
    int a, b, c, d, id;
} Q[N * 10];
int fa[N], rt[N], ans[5 * N];
bool cmp_ASK(ASK A, ASK B) {
    return A.c == B.c ? A.d < B.d : A.c < B.c; 
}
int gf(int x) {
    return x == fa[x] ? fa[x] : fa[x] = gf(fa[x]); 
}
signed main() {
    read(n); read(m); read(q); 
    for (int i = 1; i <= n; i ++) read(val[i]), id[i] = val[i], fa[i] = i; 
    sort(id + 1, id + 1 + n);
    for (int i = 1; i <= n; i ++) 
        val[i] = lower_bound(id + 1, id + 1 + n, val[i]) - id;
    for (int i = 1; i <= m; i ++) 
        read(Q[i].a), read(Q[i].b), read(Q[i].c), Q[i].d = 0;
    for (int i = m + 1; i <= m + q; i ++) 
        read(Q[i].a), read(Q[i].c), read(Q[i].b), Q[i].d = 1, Q[i].id = i - m;
    sort(Q + 1, Q + 1 + m + q, cmp_ASK);
    for (int i = 1; i <= n; i ++) seg::upd(rt[i], 1, n, val[i]);
    for (int i = 1; i <= m + q; i ++) {
        if (!Q[i].d) {
            int x = gf(Q[i].a), y = gf(Q[i].b);
            if (x != y) {
                fa[x] = y; 
                rt[y] = seg::merge(rt[x], rt[y]);
            } 
        } else {
            int x = gf(Q[i].a);
            if (seg::tr[rt[x]].s < Q[i].b) ans[Q[i].id] = -1;
            else ans[Q[i].id] = id[seg::kth(rt[x], 1, n, seg::tr[rt[x]].s - Q[i].b + 1)]; 
        }
    }
    for (int i = 1; i <= q; i ++) printf("%d\n", ans[i]);
    return 0;
}
posted @ 2019-04-28 20:59 chhokmah 阅读(...) 评论(...) 编辑 收藏