建顆權(quán)值線段樹就行了...連離散化都不用...

沒加讀入優(yōu)化就TLE, 加了就A掉了...而且還快了接近1/4....?

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#include<bits/stdc++.h>#define rep(i, n) for(int i = 0; i < n; i++)#define clr(x, c) memset(x, c, sizeof(x))#define M(l, r) (((l) + (r)) >> 1)using namespace std;const int maxn = 1000009;?struct Node {Node *l, *r;bool s;Node() {s = false;}inline void update() {s = l->s || r->s;}} pool[maxn << 1], *pt = pool, *root;void build(Node* t, int l, int r) {if(r > l) {int m = M(l, r);build(t->l = pt++, l, m);build(t->r = pt++, m + 1, r);}}int n, v, type;void modify(Node* t, int l, int r) {if(l == r) {if(type == 1 && !t->s) t->s = true;if(type == 2 && t->s) t->s = false;} else {int m = M(l, r);v <= m ? modify(t->l, l, m) : modify(t->r, m + 1, r);t->update();}}int query(Node* t, int l, int r) {if(!t->s) return 0;if(l == r) return l;int m = M(l, r);return type == 3 ? (t->l->s ? query(t->l, l, m) : query(t->r, m + 1, r)) :? ? ? ? ? ? ? ? ? ? (t->r->s ? query(t->r, m + 1, r) : query(t->l, l, m));}int succ(Node* t, int l, int r) {if(!t->s) return 0;if(l == r) return l <= v ? 0 : l;int m = M(l, r);if(v >= m) return succ(t->r, m + 1, r);int ans = succ(t->l, l, m);return ans ? ans : succ(t->r, m + 1, r);}int pred(Node* t, int l, int r) {if(!t->s) return 0;if(l == r) return l >= v ? 0 : l;int m = M(l, r);if(m + 1 >= v) return pred(t->l, l, m);int ans = pred(t->r, m + 1, r);return ans ? ans : pred(t->l, l, m);}int find(Node* t, int l, int r) {if(!t->s) return -1;if(l == r) return 1;int m = M(l, r);return v <= m ? find(t->l, l, m) : find(t->r, m + 1, r);}inline void read(int &t) {t = 0;char c = getchar();for(; !isdigit(c); c = getchar());for(; isdigit(c); c = getchar()) ? ?t = t * 10 + c - '0';}int main() {freopen("test.in", "r", stdin);freopen("test.out", "w", stdout);int m;cin >> n >> m;n++;build(root = pt++, 1, n);while(m--) {read(type);switch(type) {case 1 : read(v); v++; modify(root, 1, n); break;case 2 : read(v); v++; modify(root, 1, n); break;case 3 : printf("%d\n", query(root, 1, n) - 1); break;case 4 : printf("%d\n", query(root, 1, n) - 1); break;case 5 : read(v); v++; printf("%d\n", pred(root, 1, n) - 1); break;case 6 : read(v); v++; printf("%d\n", succ(root, 1, n) - 1); break;case 7 : read(v); v++; printf("%d\n", find(root, 1, n)); break;default : break;}}return 0;}

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3685: 普通van Emde Boas樹

Time Limit:?9 Sec??Memory Limit:?128 MB
Submit:?616??Solved:?216
[Submit][Status][Discuss]

Description

設(shè)計數(shù)據(jù)結(jié)構(gòu)支持:
1 x? 若x不存在,插入x
2 x? 若x存在,刪除x
3??? 輸出當(dāng)前最小值,若不存在輸出-1
4??? 輸出當(dāng)前最大值,若不存在輸出-1
5 x? 輸出x的前驅(qū),若不存在輸出-1
6 x? 輸出x的后繼,若不存在輸出-1
7 x? 若x存在,輸出1,否則輸出-1

Input

第一行給出n,m 表示出現(xiàn)數(shù)的范圍和操作個數(shù)
接下來m行給出操作
n<=10^6,m<=2*10^6,0<=x<n

Output

Sample Input

10 11
1 1
1 2
1 3
7 1
7 4
2 1
3
2 3
4
5 3
6 2

Sample Output

1
-1
2
2
2
-1

HINT

Source

By Zky

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轉(zhuǎn)載于:https://www.cnblogs.com/JSZX11556/p/4666047.html

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