【BZOJ3545】【BZOJ3551】[ONTAK2010]Peaks

【题目链接】

• BZOJ3545(权限题)
• BZOJ3551

【前置技能】

• 主席树
• Kruskal重构树

【题解】

• 两题其实分别是支持离线的版本和强制在线的版本。离线版本其实也可以离线询问，按困难值排序然后数据结构合并一下就好了。
• 仔细分析一下，其实有效的路径只有最小生成树上的边，因为如果走其他不在最小生成树上的路径到达某个节点，困难度的下限一定不会更小。在Kruskal重构树上倍增往上跳到最远能跳的点，那么所有能到达的点都在这个点的子树中。按照dfs序建立主席树，在主席树上二分即可。
• 时间复杂度$O(NlogN+QlogN)$

【代码】

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
#define LL  long long
#define MAXN    100010
#define MAXM    500010
#define MAXLOG  21
using namespace std;
int n, m, Q, h[MAXN], san[MAXN], v, lim, k, lastans;
struct edg{int u, v, w;}e[MAXM];
bool cmp(edg a, edg b){return a.w < b.w;}

template <typename T> void chkmax(T &x, T y){x = max(x, y);}
template <typename T> void chkmin(T &x, T y){x = min(x, y);}
template <typename T> void read(T &x){
    x = 0; int f = 1; char ch = getchar();
    while (!isdigit(ch)) {if (ch == '-') f = -1; ch = getchar();}
    while (isdigit(ch)) {x = x * 10 + ch - '0'; ch = getchar();}
    x *= f;
}

struct Segment_Tree{
    struct info{int ls, rs, sum;}a[MAXN * MAXLOG + MAXN];
    int n, root[MAXN], cnt;
    void build(int &pos, int l, int r){
        if (l == r) return;
        int mid = (l + r) >> 1;
        build(a[pos].ls, l, mid);
        build(a[pos].rs, mid + 1, r);
    }
    void init(int x){
        n = x, cnt = 0;
        build(root[0], 1, n);
    }
    void modify(int &pos, int old, int l, int r, int p){
        pos = ++cnt;
        a[pos] = a[old], ++a[pos].sum;
        if (l == r) return;
        int mid = (l + r) >> 1;
        if (p <= mid) modify(a[pos].ls, a[old].ls, l, mid, p);
        else modify(a[pos].rs, a[old].rs, mid + 1, r, p);
    }
    void modify(int Tcur, int Told, int p){
        modify(root[Tcur], root[Told], 1, n, p);
    }
    int query(int lp, int rp, int l, int r, int k){
        if (l == r) return l;
        int mid = (l + r) >> 1, tmp = a[a[rp].ls].sum - a[a[lp].ls].sum;
        if (tmp >= k) return query(a[lp].ls, a[rp].ls, l, mid, k);
        else return query(a[lp].rs, a[rp].rs, mid + 1, r, k - tmp);
    }
    int query(int L, int R, int k){
        if (a[root[R]].sum - a[root[L - 1]].sum < k) return -1;
        else return query(root[L - 1], root[R], 1, n , a[root[R]].sum - a[root[L - 1]].sum - k + 1);
    }
}sgt;

struct Kruskal_Tree{
    struct info{int ls, rs, val, L, R;}a[MAXN * 2];
    int cnt, fa[MAXN * 2][MAXLOG + 2], rnk, F[MAXN * 2];
    void dfs(int pos, int dad){
        fa[pos][0] = dad, a[pos].L = INF, a[pos].R = -INF;
        for (int i = 1; i <= MAXLOG; ++i)
            fa[pos][i] = fa[fa[pos][i - 1]][i - 1];
        if (a[pos].ls) {
            dfs(a[pos].ls, pos);
            chkmin(a[pos].L, a[a[pos].ls].L);
            chkmax(a[pos].R, a[a[pos].ls].R);
        }
        if (a[pos].rs) {
            dfs(a[pos].rs, pos);
            chkmin(a[pos].L, a[a[pos].rs].L);
            chkmax(a[pos].R, a[a[pos].rs].R);
        }
        if (pos <= n)    {
            a[pos].L = a[pos].R = ++rnk;
            sgt.modify(rnk, rnk - 1, h[pos]);
        }
    }
    int find(int x){
        if (F[x] == x) return x;
        else return (F[x] = find(F[x]));
    }
    void init(){
        cnt = n, rnk = 0;
        sort(e + 1, e + 1 + m, cmp);
        for (int i = 1; i <= 2 * n; ++i)
            F[i] = i;
        int tmp = 0;
        for (int i = 1; i <= m; ++i){
            int x = e[i].u, y = e[i].v, w = e[i].w, fx = find(x), fy = find(y);
            if (fx != fy){
                ++cnt;
                a[cnt].ls = fx, a[cnt].rs = fy, a[cnt].val = w;
                F[fx] = F[fy] = cnt;
                ++tmp;
                if (tmp == n - 1) break;
            }
        }
        dfs(cnt, 0);
    }
    int query(int v, int lim, int k){
        for (int i = MAXLOG; i >= 0; --i)
            if (fa[v][i] && a[fa[v][i]].val <= lim) v = fa[v][i];
        return sgt.query(a[v].L, a[v].R, k);
    }
}kt;

void discre(){
    for (int i = 1; i <= n; ++i)
        san[i] = h[i];
    sort(san + 1, san + 1 + n);
    int cnt = unique(san + 1, san + 1 + n) - san - 1;
    for (int i = 1; i <= n; ++i)
        h[i] = lower_bound(san + 1, san + 1 + cnt, h[i]) - san;
}

int main(){
    read(n), read(m), read(Q);
    for (int i = 1; i <= n; ++i)
        read(h[i]);
    discre();
    for (int i = 1; i <= m; ++i)
        read(e[i].u), read(e[i].v), read(e[i].w);
    sgt.init(n);
    kt.init();
    while (Q--){
        read(v), read(lim), read(k);
        if (lastans != -1){
            v = v ^ lastans, lim = lim ^ lastans, k = k ^ lastans;
        }
        lastans = kt.query(v, lim, k);
        if (lastans != -1) lastans = san[lastans];
        printf("%d\n", lastans);
    }
    return 0;
}