https://www.luogu.com.cn/problem/P3369

treap和bst（二叉查找樹(shù)）的區(qū)別就在于 treap 給了每個(gè)節(jié)點(diǎn)一個(gè)隨機(jī)的權(quán)值，在插入的時(shí)候，通過(guò)該權(quán)值使整棵樹(shù)依然維持一個(gè)大根堆/小根堆的形式（使用左旋和右旋方式來(lái)完成），在刪除的時(shí)候，通過(guò)將該點(diǎn)旋轉(zhuǎn)到葉子結(jié)點(diǎn)后再刪除。

代碼參考 <算法競(jìng)賽進(jìn)階指南>

//https://www.luogu.com.cn/problem/P3369 //treap #include <bits/stdc++.h> #define ll long long #define sc scanf #define pr printf using namespace std; const int MAXN = 1e5 + 5; const ll inf = 1e9 + 7; struct node {int val;//權(quán)值int sz;//子樹(shù)大小int ch[2];//左右兒子int cnt;//該權(quán)值個(gè)數(shù)int data;//隨機(jī)數(shù)（用來(lái)平衡二叉樹(shù)） }treap[MAXN]; int cnt, root; int newnode(int val) {treap[cnt].val = val;treap[cnt].sz = 1;treap[cnt].ch[0] = treap[cnt].ch[1] = 0;treap[cnt].cnt = 1;treap[cnt].data = rand();return cnt++; } void up(int rot) {treap[rot].sz = treap[treap[rot].ch[0]].sz + treap[treap[rot].ch[1]].sz + treap[rot].cnt; } void build() {cnt = 1;newnode(-inf); newnode(inf);root = 1;treap[1].ch[1] = 2;up(1); } void rotate(int& rot, int op) {int t = treap[rot].ch[op ^ 1];treap[rot].ch[op ^ 1] = treap[t].ch[op];treap[t].ch[op] = rot;rot = t;up(treap[rot].ch[op]);up(rot); } void insert(int& rot, int val) {if (rot == 0){rot = newnode(val);return;}if (treap[rot].val > val){insert(treap[rot].ch[0], val);if (treap[rot].data < treap[treap[rot].ch[0]].data)rotate(rot, 1);}else if (treap[rot].val < val){insert(treap[rot].ch[1], val);if (treap[rot].data < treap[treap[rot].ch[1]].data)rotate(rot, 0);}elsetreap[rot].cnt++;up(rot); } int find(int rot, int val) {if (rot == 0)return -1;if (treap[rot].val == val)return rot;else if (treap[rot].val < val)return find(treap[rot].ch[1], val);elsereturn find(treap[rot].ch[0], val); } int getpre(int rot, int val)//求比val小的最大的數(shù)字 {if (rot == 0)return 1;//最小值int ans = 1;//答案在bst中的下標(biāo)if (treap[rot].val < val)ans = rot;if (treap[rot].val < val){int t = getpre(treap[rot].ch[1], val);if (treap[ans].val < treap[t].val)ans = t;}else{int t = getpre(treap[rot].ch[0], val);if (treap[ans].val < treap[t].val)ans = t;}return ans; } int getnex(int rot, int val)//求比val大的最小的數(shù)字 {if (rot == 0)return 2;//最大值int ans = 2;if (treap[rot].val > val)ans = rot;if (treap[rot].val <= val){int t = getnex(treap[rot].ch[1], val);if (treap[ans].val > treap[t].val)ans = t;}else{int t = getnex(treap[rot].ch[0], val);if (treap[ans].val > treap[t].val)ans = t;}return ans; } void delet(int& rot, int val) {if (rot == 0)return;if (treap[rot].val == val){if (treap[rot].cnt != 1)treap[rot].cnt--;else if (treap[rot].ch[0] || treap[rot].ch[1]){//判斷左旋還是右旋，將需要?jiǎng)h除的節(jié)點(diǎn)旋轉(zhuǎn)到葉子節(jié)點(diǎn)if (treap[rot].ch[1] == 0 || treap[treap[rot].ch[0]].data > treap[treap[rot].ch[1]].data){rotate(rot, 1);delet(treap[rot].ch[1], val);}else{rotate(rot, 0);delet(treap[rot].ch[0], val);}}elserot = 0;}else if (treap[rot].val > val)delet(treap[rot].ch[0], val);elsedelet(treap[rot].ch[1], val);up(rot); } int numrank(int rot, int val) {if (rot == 0)return 0;if (treap[rot].val == val)return treap[treap[rot].ch[0]].sz + 1;else if (treap[rot].val > val)return numrank(treap[rot].ch[0], val);elsereturn treap[treap[rot].ch[0]].sz + treap[rot].cnt + numrank(treap[rot].ch[1], val); } int kthnum(int rot, int k) {if (k <= treap[treap[rot].ch[0]].sz)return kthnum(treap[rot].ch[0], k);else if (k > treap[treap[rot].ch[0]].sz + treap[rot].cnt)return kthnum(treap[rot].ch[1], k - (treap[treap[rot].ch[0]].sz + treap[rot].cnt));elsereturn rot; } int main() {int n;sc("%d", &n);build();while (n--){int op, a;sc("%d%d", &op, &a);switch (op){case 1:insert(root, a); break;case 2:delet(root, a); break;case 3:pr("%d\n", numrank(root, a) - 1); break;case 4:pr("%d\n", treap[kthnum(root, a + 1)].val); break;case 5:pr("%d\n", treap[getpre(root, a)].val); break;case 6:pr("%d\n", treap[getnex(root, a)].val); break;}} }

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