BZOJ 2733: [HNOI2012]永无乡启发式合并treap

2733: [HNOI2012]永无乡

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://www.lydsy.com/JudgeOnline/problem.php?id=2733

Description

永无乡包含 n 座岛，编号从 1 到 n，每座岛都有自己的独一无二的重要度，按照重要度可以将这 n 座岛排名，名次用 1 到 n 来表示。某些岛之间由巨大的桥连接，通过桥可以从一个岛到达另一个岛。如果从岛 a 出发经过若干座（含 0 座）桥可以到达岛 b，则称岛 a 和岛 b 是连通的。现在有两种操作：B x y 表示在岛 x 与岛 y 之间修建一座新桥。Q x k 表示询问当前与岛 x连通的所有岛中第 k 重要的是哪座岛，即所有与岛 x 连通的岛中重要度排名第 k 小的岛是哪座，请你输出那个岛的编号。

Input

输入文件第一行是用空格隔开的两个正整数 n 和 m，分别表示岛的个数以及一开始存在的桥数。接下来的一行是用空格隔开的 n 个数，依次描述从岛 1 到岛 n 的重要度排名。随后的 m 行每行是用空格隔开的两个正整数 ai 和 bi，表示一开始就存在一座连接岛 ai 和岛 bi 的桥。后面剩下的部分描述操作，该部分的第一行是一个正整数 q，表示一共有 q 个操作，接下来的 q 行依次描述每个操作，操作的格式如上所述，以大写字母 Q 或B 开始，后面跟两个不超过 n 的正整数，字母与数字以及两个数字之间用空格隔开。对于 20%的数据 n≤1000,q≤1000

对于 100%的数据 n≤100000,m≤n，q≤300000

Output

对于每个 Q x k 操作都要依次输出一行，其中包含一个整数，表示所询问岛屿的编号。如果该岛屿不存在，则输出-1。

Sample Input

Sample Output

HINT

题意

题解：

启发式合并treap，直接套版咯。。。

合并就用并查集来维护，如果不在的话，就强行把小的treap直接暴力insert到大的treap里面就好了

然后再找第k大就好了

代码

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<vector>
#include<algorithm>
#include<ctime>
using namespace std;#define maxn 100005
int n,m,q;
//
struct TreeNode {TreeNode *L, *R;int num;int size;int pri;int key;
};
TreeNode nodes[maxn*30], stack[maxn*5];
TreeNode *null;
int C, Top;void init() {C = 0; Top = 0;null = &nodes[C++];null->L = null->R = null;null->pri = -0x7FFFFFFF;null->size = 0;null->key = 0;null->num = 0;
}struct Treap {TreeNode* root;vector<TreeNode*> list;void init() {root = null;}void toArrayList(TreeNode* root) {if (root != null) {toArrayList(root->L);list.push_back(root);toArrayList(root->R);}}TreeNode* newNode(int key, int num) {TreeNode* ret;if (Top)ret = &stack[--Top];else ret = &nodes[++C];ret->L = ret->R = null;ret->key = key;ret->pri = rand();ret->num = num;ret->size = num;return ret;}TreeNode* merge(TreeNode* A , TreeNode* B) {//合并if (A == null)return B;if (B == null)return A;if (A->pri < B->pri) {A->R = merge(A->R, B);pushUp(A);return A;} else {B->L = merge(A, B->L);pushUp(B);return B;}}pair<TreeNode*, TreeNode*> split(TreeNode* root, int key) {//分裂pair<TreeNode*, TreeNode*> tmp;if (root == null) {tmp = make_pair(null, null);return tmp;}if (key <= root->key) {tmp = split(root->L, key);root->L = tmp.second;pushUp(root);tmp.second = root;return tmp;} else {tmp = split(root->R, key);root->R = tmp.first;pushUp(root);tmp.first = root;return tmp;}}TreeNode* upperBound(TreeNode* root, int key) {TreeNode* ans = null;TreeNode* i = root;while (i != null) {if (key < i->key) {ans = i;i = i->L;} else {i = i->R;}}return ans;}TreeNode* lowerBound(TreeNode* root, int key) {TreeNode* ans = null;TreeNode* i = root;while (i != null) {if (key < i->key) {i = i->L;} else {ans = i;i = i->R;}}return ans;}bool contains(TreeNode* root, int key) {if (root == null)return false;if (key == root->key)return true;else if (key < root->key)return contains(root->L, key);elsereturn contains(root->R, key);}void insert(TreeNode* root, int key, int num) {if (key < root->key)insert(root->L, key, num);else if (key == root->key)root->num += num;elseinsert(root->R, key, num);pushUp(root);}void insert(int key, int num) {if (contains(root, key)) {insert(root, key, num);return ;}pair<TreeNode*, TreeNode*> tmp = split(root, key);TreeNode* node = newNode(key, num);root = merge(merge(tmp.first, node), tmp.second);}void del(int key) {pair<TreeNode*, TreeNode*> tmp = split(root, key);TreeNode* node = upperBound(tmp.second, key);if (node == null)root = tmp.first;elseroot = merge(tmp.first, split(tmp.second, node->key).second);}void del(TreeNode* root, int key) {if (!contains(root, key))return ;if (key < root->key)del(root->L, key);else if (key == root->key) {root->num--;if (root->num == 0)del(key);} else {del(root->R, key);}pushUp(root);}TreeNode* select(TreeNode* root, int k) {if (k <= root->L->size)return select(root->L, k);else if (k >= root->L->size + 1 && k <= root->L->size + root->num)return root;elsereturn select(root->R, k-root->L->size-root->num);}int getMin(TreeNode* root, int key) {if (key < root->key)return getMin(root->L, key);else if (key == root->key)return root->L->size + root->num;elsereturn root->L->size + root->num + getMin(root->R, key);}int getMax(TreeNode* root, int key) {if (key < root->key)return root->R->size + root->num + getMax(root->L, key);else if (key == root->key)return root->R->size + root->num;elsereturn getMax(root->R, key);}int size() {return root->size;}void visit(TreeNode* root) {if (root != null) {printf("key: %d num: %d size: %d\n", root->key, root->num, root->size);visit(root->L);visit(root->R);}}void pushUp(TreeNode* x) {x->size = x->L->size + x->R->size + x->num;}
};//
int fa[maxn];
Treap tr[maxn];
int val[maxn],Hash[maxn];
int fi(int x)
{if(x!=fa[x])fa[x]=fi(fa[x]);return fa[x];
}
void uni(int x,int y)
{x = fi(x),y = fi(y);if(x!=y){if(tr[x].size()>tr[y].size())swap(x,y);fa[x]=y;tr[x].list.clear();tr[x].toArrayList(tr[x].root);for(int i=0;i<tr[x].list.size();i++){TreeNode* tmp = tr[x].list[i];tr[y].insert(tmp->key,tmp->num);}}
}
int query(int x,int k)
{x = fi(x);if(tr[x].size()<k)return -1;else return tr[x].select(tr[x].root,k)->key;
}
int main()
{init();scanf("%d%d",&n,&m);for(int i=1;i<=n;i++){tr[i].init();fa[i]=i;scanf("%d",&val[i]);Hash[val[i]]=i;tr[i].insert(val[i],1);}for(int i=1;i<=m;i++){int x,y;scanf("%d%d",&x,&y);uni(x,y);}scanf("%d",&q);for(int i=0;i<q;i++){char s[10];scanf("%s",&s);int x,y;scanf("%d%d",&x,&y);if(s[0]=='B'){uni(x,y);}else{int p = query(x,y);if(p==-1)printf("-1\n");else printf("%d\n",Hash[p]);}}
}

转载于:https://www.cnblogs.com/qscqesze/p/4969342.html