SPOJ 4487. Can you answer these queries VI splay

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题目链接：点击打开链接

题意比较明显，不赘述。

删除时可以把i-1转到根，把i+1转到根下

则i点就在 根右子树 的左子树，且只有i这一个 点

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 300500
#define inf 10000000
#define L(x) tree[x].ch[0]
#define R(x) tree[x].ch[1]
#define Father(x) tree[x].fa
#define Size(x) tree[x].size
#define Val(x) tree[x].val
#define Left(x) tree[x].left
#define Right(x) tree[x].right
#define Sum(x) tree[x].sum
#define Ans(x) tree[x].ans
struct node{
	int ch[2],fa,size;
	int val, left, right, sum, ans;
}tree[N];
int tot, root;
int a[N];
void Newnode(int& id, int val, int fa){
	node E={0,0,fa,1,val,val,val,val,val};
	id = tot++;
	tree[id] = E;
}
void push_up(int id){
	Size(id) = Size(L(id))+Size(R(id))+1;
	Sum(id) = Sum(L(id)) + Sum(R(id)) + Val(id);
	Left(id) = max(Left(L(id)), Sum(L(id))+Val(id)+max(Left(R(id)), 0));
	Right(id) = max(Right(R(id)), Sum(R(id))+Val(id)+max(Right(L(id)), 0));
	Ans(id) = max(Ans(L(id)), Ans(R(id)));
	Ans(id) = max(Ans(id), Val(id)+max(Right(L(id)),0)+max(Left(R(id)),0));
}
void push_down(int id){}
void Rotate(int id, int kind){
	int y = Father(id);
	push_down(id); push_down(y);
	tree[y].ch[kind^1] = tree[id].ch[kind];
	Father(tree[id].ch[kind]) = y;
	if(Father(y))
		tree[Father(y)].ch[R(Father(y))==y] = id;
	Father(id) = Father(y);
	Father(y) = id;
	tree[id].ch[kind] = y;
	push_up(y);
}
void Splay(int id, int goal){
	push_down(id);
	while(Father(id)!=goal){
		int y = Father(id);
		if(Father(y)==goal)
			Rotate(id,L(y)==id);
		else {
			int kind = L(Father(y))==y;
			if(tree[y].ch[kind]==id)
			{
				Rotate(id, kind^1);
				Rotate(id, kind);
			}
			else 
			{
				Rotate(y, kind);
				Rotate(id, kind);
			}
		}
		push_down(id);
	}
	if(goal==0)root = id;
	push_up(id);
}
int Get_kth(int k){
	int id = root;
	push_down(id);
	while(Size(L(id))!=k){
		if(Size(L(id))>k)
			id = L(id);
		else {
			k -= (Size(L(id)) +1);
			id = R(id);
		}
		push_down(id);
	}
	return id;
}

int Getmax(int id){
	push_down(id);
	while(R(id)){
		id = R(id);
		push_down(id);
	}
	return id;
}
void Delete(int id){
	int a = Get_kth(id-1);
	int b = Get_kth(id+1);
	Splay(a, 0);
	Splay(b, root);
	L(b) = 0;
	push_up(b);
	push_up(a);
}
int build(int l, int r, int& id, int fa){
	if(l>r)return 0;
	int mid = (l+r)>>1;
	Newnode(id, a[mid], fa);
	build(l, mid-1, L(id), id);
	build(mid+1, r, R(id), id);
	push_up(id);
}
int n, que;
char s[2];
void init(){
	Father(0) = L(0) = R(0) = Size(0) = 0;
	Sum(0) = 0;
	Val(0) = Left(0) = Right(0) = Ans(0) = -inf;
	root = tot = 1;
	Newnode(root, -inf, 0);
	Newnode(R(root), -inf, root);
	build(1, n, L(R(root)), R(root));
	push_up(R(root)); push_up(root);
}

int main(){
	int i, j;
	while(~scanf("%d",&n)){
		for(i = 1; i <= n; i++)scanf("%d",&a[i]);
		init();
		scanf("%d",&que);
		while(que--){
			scanf("%s",s);
			if(s[0]==‘I‘)
			{
				scanf("%d %d",&i,&j);
				Splay(Get_kth(i), 0);
				Splay(Getmax(L(root)), root);
				Newnode(R(L(root)), j, L(root));
				push_up(L(root));
				push_up(root);
			}
			else if(s[0]==‘Q‘)
			{
				scanf("%d %d",&i,&j);
				Splay(Get_kth(i-1), 0);
				Splay(Get_kth(j+1), root);
				printf("%d\n", Ans(L(R(root))));
			}
			else if(s[0]==‘D‘)
			{
				scanf("%d",&i);
				Delete(i);
			}
			else if(s[0]==‘R‘)
			{
				scanf("%d %d",&i,&j);
				Splay(Get_kth(i), 0);
				Val(root) = j;
				push_up(root);
			}
		}
	}
	return 0;
}
/*
5
3 -4 3 -1 6
99
I 6 2
Q 3 5
R 5 -4
Q 3 5
D 2
Q 1 5
I 2 -10
Q 1 6
R 2 -1
Q 1 6
Q 1 6
*/

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SPOJ 4487. Can you answer these queries VI splay

标签：blog   class   code   tar   c   http   

原文地址：http://blog.csdn.net/acmmmm/article/details/25535325