bzoj3545

线段树合并+离线+启发式合并

半年前这道题t成狗。。。

离线的做法比较好想，按照边的权值排序，询问的权值排序，然后枚举询问不断加边，加到上限后查找第k大值，这里平衡树，权值线段树都可以实现。

那么我们用权值线段树就行了， 并查集维护两点连通性，不连通的话就合并，并查集连接。

#include<bits/stdc++.h>
using namespace std;
const int N = 500010;
struct Query {
    int v, x, k, id;
    bool friend operator < (Query A, Query B)
    {
        return A.x < B.x;
    }
} q[N];
struct edge {
    int u, v, w;
    bool friend operator < (edge A, edge B)
    {
        return A.w < B.w;
    }
} e[N]; 
int n, m, Q;
int root[N], h[N], ans[N];
vector<int> vt;
map<int, int> mp;
inline int read()
{
    int x = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
    while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
    return x * f;
}
namespace seg 
{
    int cnt;
    int lc[N << 2], rc[N << 2], size[N << 2];
    void update(int l, int r, int &x, int pos)
    {
        x = ++cnt;
        ++size[x];
        if(l == r) return;
        int mid = (l + r) >> 1;
        if(pos <= mid) update(l, mid, lc[x], pos);
        else update(mid + 1, r, rc[x], pos);
    }
    int merge(int x, int y)
    {
        if(!x) return y;
        if(!y) return x;
        size[x] += size[y];
        lc[x] = merge(lc[x], lc[y]);
        rc[x] = merge(rc[x], rc[y]);
        return x;
    }
    int query(int l, int r, int x, int rank)
    {
        if(l == r) return l;
        int mid = (l + r) >> 1;
        if(size[rc[x]] >= rank) return query(mid + 1, r, rc[x], rank);
        else return query(l, mid, lc[x], rank - size[rc[x]]);
    }
} using namespace seg;
int main()
{
    n = read();
    m = read();
    Q = read();
    for(int i = 1; i <= n; ++i) 
    {
        h[i] = read();
        vt.push_back(h[i]);
    }
    sort(vt.begin(), vt.end());
    vt.erase(unique(vt.begin(), vt.end()), vt.end());
    for(int i = 1; i <= n; ++i) 
    {
        h[i] = lower_bound(vt.begin(), vt.end(), h[i]) - vt.begin() + 1;
        update(1, n, root[i], h[i]);
    }
    for(int i = 1; i <= m; ++i)
    {
        e[i].u = read();
        e[i].v = read();
        e[i].w = read();
    }
    sort(e + 1, e + m + 1);
    for(int i = 1; i <= Q; ++i)
    {
        q[i].v = read();
        q[i].x = read();
        q[i].k = read();
        q[i].id = i;
    }
    sort(q + 1, q + Q + 1);
    int j = 1;
    for(int i = 1; i <= Q; ++i)
    {
        while(e[j].w <= q[i].x && j <= m) 
        {
            if(root[e[j].u] != root[e[j].v]) root[e[j].u] = root[e[j].v] = merge(root[e[j].u], root[e[j].v]);
            ++j;
        }
        if(size[root[q[i].v]] < q[i].k) ans[q[i].id] = -1;
        else ans[q[i].id] = query(1, n, root[q[i].v], q[i].k);      
    }
    for(int i = 1; i <= Q; ++i) printf("%d
", ans[i]);
    return 0;
}