Problem
传送门
给你一张n个点，m条边的图，边有边权，点有点权。
现有Q个询问，每次询问从点x开始，只走边权小于y的边，能走到的点中点权第k大的点
Solution
这道题没有强制在线，所以直接想了一个简单的离线做法。
将询问按照y排序，边按照边权排序。
将边按顺序一条一条加入，合并了两个联通块的同时合并权值线段树
至于查询，在权值线段树查询全局的k大就可以了。
不知道为什么大家都打的启发式合并+主席树
Code
#include <bits/stdc++.h>
using namespace std;
#define DEBUG(...) fprintf(stderr, __VA_ARGS__)
#define mp make_pair
#define fst first
#define snd second
template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }
inline int read(){
int res = 0, fl = 1;
char r = getchar();
for (; !isdigit(r); r = getchar()) if(r == '-') fl = -1;
for (; isdigit(r); r = getchar()) res = (res << 3) + (res << 1) + r - 48;
return res * fl;
}
typedef long long LL;
typedef pair<int, int> pii;
const int Maxn = 5e5 + 10;
namespace SGT{
int tre[Maxn << 4], ls[Maxn << 4], rs[Maxn << 4], cnt;
#define mid ((l + r) >> 1)
#define sum tre[rs[rt]]
inline void Insert(int &rt, int l, int r, int pos){
if(!rt) rt = ++cnt;
tre[rt]++;
if(l == r) return;
if(mid >= pos) Insert(ls[rt], l, mid, pos);
else Insert(rs[rt], mid + 1, r, pos);
}
inline int Query(int rt ,int l, int r, int k){
if(tre[rt] < 0) return -1;
if(tre[rt] < k) return -1;
if(l == r) return l;
if(sum >= k) return Query(rs[rt], mid + 1, r, k);
else return Query(ls[rt], l, mid, k - sum);
}
inline int merge(int u, int v){
if(!u) return v;
if(!v) return u;
tre[u] += tre[v];
ls[u] = merge(ls[u], ls[v]);
rs[u] = merge(rs[u], rs[v]);
return u;
}
#undef mid
#undef sum
}
struct node {
int x, y, val, id;
bool operator < (const node T) const{ return val < T.val;}
}g[Maxn],ASK[Maxn];
int ans[Maxn], fa[Maxn], h[Maxn], b[Maxn], root[Maxn], num;
bool vis[Maxn];
vector <int> G[Maxn];
inline int get_fa(int x){ return fa[x] == x ? x : fa[x] = get_fa(fa[x]);}
inline void link(int u, int v){
if(get_fa(u) == get_fa(v)) return;
root[get_fa(v)] = SGT::merge(root[get_fa(u)], root[get_fa(v)]);
fa[get_fa(u)] = get_fa(v);
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("a.in", "r", stdin);
freopen("a.out", "w", stdout);
#endif
int n = read(), m = read(), Q = read();
for (int i = 1; i <= n; ++i) b[i] = h[i] = read();
sort(b + 1, b + 1 + n);
num = unique(b + 1, b + 1 + n) - b - 1;
for (int i = 1; i <= n; ++i) h[i] = lower_bound(b + 1, b + 1 + num, h[i]) - b;
for (int i = 1; i <= m; ++i)
g[i].x = read(), g[i].y = read(), g[i].val = read();
sort(g + 1, g + 1 + m);
for (int i = 1; i <= Q; ++i)
ASK[i].x = read(), ASK[i].val = read(), ASK[i].y = read(),ASK[i].id = i;
sort(ASK + 1, ASK + 1 + Q);
for (int i = 1; i <= n; ++i) fa[i] = i;
for (int i = 1; i <= n; ++i)
SGT::Insert(root[i], 1, num, h[i]);
int top = 1;
for (int i = 1; i <= m; ++i){
for(; ASK[top].val < g[i].val && top <= Q; ++top){
ans[ASK[top].id] = SGT::Query(root[get_fa(ASK[top].x)], 1, num, ASK[top].y);
vis[ASK[top].id] = 1;
}
link(g[i].x, g[i].y);
}
for (int i = top; i <= Q; ++i)
ans[ASK[i].id] = SGT::Query(root[get_fa(ASK[i].x)], 1, num, ASK[i].y);
b[0] = -1;
for (int i = 1; i <= Q; ++i)
printf("%d\n",b[max(ans[i], 0)]);
return 0;
}
转载于:https://www.cnblogs.com/LZYcaiji/p/10611169.html
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