[BZOJ3545][ONTAK2010]Peaks
试题描述
在Bytemountains有N座山峰，每座山峰有他的高度h_i。有些山峰之间有双向道路相连，共M条路径，每条路径有一个困难值，这个值越大表示越难走，现在有Q组询问，每组询问询问从点v开始只经过困难值小于等于x的路径所能到达的山峰中第k高的山峰，如果无解输出-1。
输入
第一行三个数N，M，Q。
第二行N个数，第i个数为h_i
接下来M行，每行3个数a b c，表示从a到b有一条困难值为c的双向路径。
接下来Q行，每行三个数v x k，表示一组询问。
第二行N个数，第i个数为h_i
接下来M行，每行3个数a b c，表示从a到b有一条困难值为c的双向路径。
接下来Q行，每行三个数v x k，表示一组询问。
输出
对于每组询问，输出一个整数表示答案。
输入示例
10 11 4 1 2 3 4 5 6 7 8 9 10 1 4 4 2 5 3 9 8 2 7 8 10 7 1 4 6 7 1 6 4 8 2 1 5 10 8 10 3 4 7 3 4 6 1 5 2 1 5 6 1 5 8 8 9 2
输出示例
6 1 -1 8
数据规模及约定
N<=10^5, M,Q<=5*10^5，h_i,c,x<=10^9。
题解
考虑离线做法，我们把边和询问按权值排一遍序，然后依次处理每个询问。那么就是从小到大依次加入那些边，对于连通性，我们可以用并查集维护；对于第 k 大值，我们可以并查集里面套一个 treap；啊 treap 怎么合并？只好加一个 log 启发式合并了。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;
const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
if(Head == Tail) {
int l = fread(buffer, 1, BufferSize, stdin);
Tail = (Head = buffer) + l;
}
return *Head++;
}
int read() {
int x = 0, f = 1; char c = Getchar();
while(!isdigit(c)){ if(c == '-') f = -1; c = Getchar(); }
while(isdigit(c)){ x = x * 10 + c - '0'; c = Getchar(); }
return x * f;
}
#define maxn 100010
#define maxm 500010
struct Node {
int v, r, siz;
Node() {}
Node(int _, int __): v(_), r(__) {}
} ns[maxn];
int ToT, fa[maxn], ch[maxn][2], rec[maxn], rcnt;
int getnode() {
if(rcnt) {
int o = rec[rcnt--];
fa[o] = ch[o][0] = ch[o][1] = 0;
return o;
}
return ++ToT;
}
void maintain(int o) {
ns[o].siz = 1;
for(int i = 0; i < 2; i++) if(ch[o][i])
ns[o].siz += ns[ch[o][i]].siz;
return ;
}
void rotate(int u) {
int y = fa[u], z = fa[y], l = 0, r = 1;
if(z) ch[z][ch[z][1]==y] = u;
if(ch[y][1] == u) swap(l, r);
fa[u] = z; fa[y] = u; fa[ch[u][r]] = y;
ch[y][l] = ch[u][r]; ch[u][r] = y;
maintain(y); maintain(u);
return ;
}
void insert(int& o, int v) {
if(!o) {
ns[o = getnode()] = Node(v, rand());
return maintain(o);
}
bool d = v > ns[o].v;
insert(ch[o][d], v); fa[ch[o][d]] = o;
if(ns[ch[o][d]].r > ns[o].r) {
int t = ch[o][d];
rotate(t); o = t;
}
return maintain(o);
}
int val[maxn], cntv;
void recycle(int& o) {
if(!o) return ;
recycle(ch[o][0]); recycle(ch[o][1]);
rec[++rcnt] = o; val[++cntv] = ns[o].v; fa[o] = 0; o = 0;
return ;
}
void merge(int& u, int& v) {
// printf("merge(%d, %d)
", u, v);
cntv = 0; recycle(v);
// printf("vals: "); for(int i = 1; i <= cntv; i++) printf("%d%c", val[i], i < cntv ? ' ' : '
');
for(int i = 1; i <= cntv; i++) insert(u, val[i]);
return ;
}
int Find(int o, int k) {
if(!o) return -1;
int rs = ch[o][1] ? ns[ch[o][1]].siz : 0;
if(k == rs + 1) return ns[o].v;
if(k < rs + 1) return Find(ch[o][1], k);
return Find(ch[o][0], k - rs - 1);
}
int pa[maxn], rt[maxn];
int findset(int x) { return x == pa[x] ? x : pa[x] = findset(pa[x]); }
struct Edge {
int a, b, c;
Edge() {}
Edge(int _1, int _2, int _3): a(_1), b(_2), c(_3) {}
bool operator < (const Edge& t) const { return c < t.c; }
} es[maxm];
struct Que {
int u, x, k, id;
Que() {}
Que(int _1, int _2, int _3, int _4): u(_1), x(_2), k(_3), id(_4) {}
bool operator < (const Que& t) const { return x < t.x; }
} qs[maxm];
int ans[maxm];
int main() {
int n = read(), m = read(), q = read();
for(int i = 1; i <= n; i++) {
int v = read();
pa[i] = i; insert(rt[i], v);
}
for(int i = 1; i <= m; i++) {
int a = read(), b = read(), c = read();
es[i] = Edge(a, b, c);
}
sort(es + 1, es + m + 1);
for(int i = 1; i <= q; i++) {
int u = read(), x = read(), k = read();
qs[i] = Que(u, x, k, i);
}
sort(qs + 1, qs + q + 1);
// for(int i = 1; i <= m; i++) printf("Edge: %d %d %d
", es[i].a, es[i].b, es[i].c);
// for(int i = 1; i <= q; i++) printf("Que: %d %d %d
", qs[i].u, qs[i].x, qs[i].k);
for(int i = 1, e = 1; i <= q; i++) {
while(e <= m && es[e].c <= qs[i].x) {
int u = findset(es[e].a), v = findset(es[e].b);
// printf("%d: %d(%d) %d(%d) %d
", e, u, es[e].a, v, es[e].b, findset(8));
if(u != v) {
if(ns[rt[u]].siz < ns[rt[v]].siz) swap(u, v);
merge(rt[u], rt[v]);
pa[v] = u;
}
e++;
}
ans[qs[i].id] = Find(rt[findset(qs[i].u)], qs[i].k);
}
for(int i = 1; i <= q; i++) printf("%d
", ans[i]);
return 0;
}
在线做法，详见这里。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;
const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
if(Head == Tail) {
int l = fread(buffer, 1, BufferSize, stdin);
Tail = (Head = buffer) + l;
}
return *Head++;
}
int read() {
int x = 0, f = 1; char c = Getchar();
while(!isdigit(c)){ if(c == '-') f = -1; c = Getchar(); }
while(isdigit(c)){ x = x * 10 + c - '0'; c = Getchar(); }
return x * f;
}
#define maxn 200010
#define maxm 500010
#define maxlog 18
#define maxnode 6666666
int n;
int ToT, sumv[maxnode], lc[maxnode], rc[maxnode], rt[maxn];
void update(int& y, int x, int l, int r, int p) {
sumv[y = ++ToT] = sumv[x] + 1;
if(l == r) return ;
int mid = l + r >> 1; lc[y] = lc[x]; rc[y] = rc[x];
if(p <= mid) update(lc[y], lc[x], l, mid, p);
else update(rc[y], rc[x], mid + 1, r, p);
return ;
}
struct Edge {
int a, b, c;
Edge() {}
Edge(int _1, int _2, int _3): a(_1), b(_2), c(_3) {}
bool operator < (const Edge& t) const { return c < t.c; }
} es[maxm];
int p_ToT, fa[maxlog][maxn], ch[maxn][2], p_val[maxn], e_val[maxn], dl[maxn], dr[maxn], clo, pid[maxn];
void build(int u) {
if(!u) return ;
// printf("%d u: %d %d %d
", e_val[u], u, ch[u][0], ch[u][1]);
dl[u] = ++clo; pid[clo] = u;
for(int i = 1; i < maxlog; i++) fa[i][u] = fa[i-1][fa[i-1][u]];
for(int i = 0; i < 2; i++) build(ch[u][i]);
dr[u] = clo;
return ;
}
int pa[maxn], id[maxn];
int findset(int x) { return x == pa[x] ? x : pa[x] = findset(pa[x]); }
int num[maxn];
int query(int u, int mx, int k, int id) {
for(int i = maxlog - 1; i >= 0; i--) if(fa[i][u] && e_val[fa[i][u]] <= mx) u = fa[i][u];
// printf("%d %d
", u, e_val[u]);
int lrt = rt[dl[u]-1], rrt = rt[dr[u]];
int l = 1, r = n;
while(l < r) {
int mid = l + r >> 1;
// printf("[%d, %d]: %d %d
", mid + 1, r, sumv[rc[rrt]] - sumv[rc[lrt]], k);
if((rrt && rc[rrt] ? sumv[rc[rrt]] : 0) - (lrt && rc[lrt] ? sumv[rc[lrt]] : 0) < k) {
k -= sumv[rc[rrt]] - sumv[rc[lrt]]; r = mid;
if(lrt) lrt = lc[lrt]; if(rrt) rrt = lc[rrt];
}
else {
l = mid + 1;
if(lrt) lrt = rc[lrt]; if(rrt) rrt = rc[rrt];
}
}
int ans = sumv[rrt] - sumv[lrt] >= k ? l : -1;
return ans < 0 ? ans : num[ans];
}
int main() {
freopen("data.in", "r", stdin);
freopen("data.out", "w", stdout);
n = read(); int m = read(), q = read();
for(int i = 1; i <= n; i++) num[i] = p_val[i] = read(), pa[i] = id[i] = i;
sort(num + 1, num + n + 1);
for(int i = 1; i <= n; i++) p_val[i] = lower_bound(num + 1, num + n + 1, p_val[i]) - num;
for(int i = 1; i <= m; i++) {
int a = read(), b = read(), c = read();
es[i] = Edge(a, b, c);
}
sort(es + 1, es + m + 1);
p_ToT = n;
for(int i = 1; i <= m; i++) {
int u = findset(es[i].a), v = findset(es[i].b);
if(u != v) {
ch[++p_ToT][0] = id[u];
ch[p_ToT][1] = id[v];
fa[0][id[v]] = fa[0][id[u]] = p_ToT;
e_val[p_ToT] = es[i].c;
pa[v] = u; id[u] = p_ToT;
}
}
build(id[findset(1)]);
for(int i = 1; i <= clo; i++)
if(p_val[pid[i]]) update(rt[i], rt[i-1], 1, n, p_val[pid[i]]);
else rt[i] = rt[i-1];
int lst = 0;
for(int i = 1; i <= q; i++) {
int v = read() ^ lst, x = read() ^ lst, k = read() ^ lst;
lst = query(v, x, k, i);
printf("%d
", lst); if(lst < 0) lst = 0;
lst = 0;
}
return 0;
}