Splay树（区间更新）—— POJ 3468 A Simple Problem with Integers

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A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 72765		Accepted: 22465
Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... , A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁, A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

Sample Output

4
55
9
15

Hint

The sums may exceed the range of 32-bit integers.

题意：

输入n，m；表示有n个数和m个询问。接着输入每个数的初始值。接着每个询问表示：Q表示求解i,j区间的和，C表示i~j区间的每个值都+k

思路：

Splay树区间更新问题，注意在更新标记后向上更新。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define MAX(x, y) ((x) > (y) ? (x) : (y))

typedef struct TREE
{
    long long sum; //以该结点为根的树的元素和
    TREE *fa, *l, *r;
    int sz; //以该结点为根的树的总结点数
	int add; //懒惰标记
}Tree;

void Push_down(Tree *T)
{
    if(NULL == T) return;
    if(T->add){
		T->sum += ((long long)(T->sz) * T->add);
        if(T->l) T->l->add += T->add;
        if(T->r) T->r->add += T->add;
        T->add = 0;
    }
}

void Init(Tree *&T, int n)
{
    int i;
	long long a;
    Tree *cur, *pre;
	pre = (Tree *)malloc(sizeof(Tree));
	pre->sum = 0;
	pre->fa = pre->l = pre->r = NULL;
	pre->sz = 1;
	pre->add = 0;
    for(i = n - 1; i > -1; i--){
        cur = (Tree *)malloc(sizeof(Tree));
		scanf("%I64d", &a);
		cur->l = pre;
		cur->r = NULL;
		pre->fa = cur;
		cur->sz = pre->sz + 1;
		cur->sum = a + pre->sum;
		cur->add = 0;
        pre = cur;
		if(0 == i){
        	cur = (Tree *)malloc(sizeof(Tree));
			cur->fa = NULL;
			cur->l = pre;
			cur->r = NULL;
			pre->fa = cur;
			cur->sz = pre->sz + 1;
			cur->sum = pre->sum;
			cur->add = 0;
		}
    }
    T = cur;
}

void R_rotate(Tree *x)
{
    Tree *y = x->fa;
    Tree *z = y->fa;
    Tree *k = x->r;
    int sx = x->sz, sy = y->sz, sk = 0;
    long long sumx = x->sum, sumy = y->sum, sumk = 0;
    if(k){
		Push_down(k);
		sk = k->sz;
		sumk = k->sum;
	}
    y->l = k;
    x->r = y;
    if(z){
        if(y == z->l) z->l = x;
        else z->r = x;
    }
    if(k) k->fa = y;
    y->fa = x;
    x->fa = z;
    y->sz = sy - sx + sk;
    x->sz = sx - sk + y->sz;
    y->sum = sumy - sumx + sumk;
    x->sum = sumx - sumk + y->sum;
}

void L_rotate(Tree *x)
{
    Tree *y = x->fa;
    Tree *z = y->fa;
    Tree *k = x->l;
    int sx = x->sz, sy = y->sz, sk = 0;
    long long sumx = x->sum, sumy = y->sum, sumk = 0;
    if(k){
		Push_down(k);
		sk = k->sz;
		sumk = k->sum;
	}
    y->r = k;
    x->l = y;
    if(z){
        if(y == z->r) z->r = x;
        else z->l = x;
    }
    if(k) k->fa = y;
    y->fa = x;
    x->fa = z;
    y->sz = sy - sx + sk;
    x->sz = sx - sk + y->sz;
    y->sum = sumy - sumx + sumk;
    x->sum = sumx - sumk + y->sum;
}

//寻找第x个数的结点
Tree *FindTag(Tree *T, int x)
{
    if(NULL == T) return NULL;
    Push_down(T);
    Tree *p;
    p = T;
    int sum = (p->l ? p->l->sz : 0) + 1;
    while(sum != x && p)
    {
        if(sum < x){
            p = p->r;
            x -= sum;
        }
        else p = p->l;
        Push_down(p);
        sum = (p->l ? p->l->sz : 0) + 1;
    }
    return p;
}

void Splay(int x, Tree *&T)
{
    Push_down(T);
    Tree *p, *X, *end, *new_t;
    end = T->fa;
    new_t = T;
    if(end) new_t = T->fa;
    X = FindTag(new_t, x);
    while(X->fa != end)
    {
        p = X->fa;
        if(end == p->fa){ //p是根结点
            if(X == p->l) R_rotate(X);
            else L_rotate(X);
            break;
        }
        //p不是根结点
        if(X == p->l){
            if(p == p->fa->l){
                R_rotate(p); //LL
                R_rotate(X); //LL
            }
            else{
                R_rotate(X); //RL
                L_rotate(X);
            }
        }
        else{
            if(p == p->fa->r){ //RR
                L_rotate(p);
                L_rotate(X);
            }
            else{ //LR
                L_rotate(X);
                R_rotate(X);
            }
        }
    }
    T = X;
}

void FreeTree(Tree *T)
{
    if(NULL == T) return;
    FreeTree(T->l);
    FreeTree(T->r);
    free(T);
}

void InOrder(Tree *T)
{
	if(NULL == T) return;
	Push_down(T);
	InOrder(T->l);
	printf("%d ", T->sum);
	InOrder(T->r);
}

int main()
{
    //freopen("in.txt", "r", stdin);
    Tree *T;
	T = NULL;
	int n, q;
	int a, b, c;
	char s[3];
	scanf("%d%d", &n, &q);
	Init(T, n);
	while(q--)
	{
		//InOrder(T);
		//printf("\n");
		scanf("%s", s);
		if('Q' == s[0]){
			scanf("%d%d", &a, &b);
			a++; b++;
			Splay(a - 1, T);
			Splay(b + 1, T->r);
			Push_down(T->r->l);
			printf("%I64d\n", T->r->l->sum);
		}
		else{
			scanf("%d%d%d", &a, &b, &c);
			a++; b++;
			Splay(a - 1, T);
			Splay(b + 1, T->r);
			T->r->l->add += c;
			//向上更新
			T->r->sum += ((long long)(b - a + 1) * c);
			T->sum += ((long long)(b - a + 1) * c);
		}
	}
	FreeTree(T);
    return 0;
}