【POJ2761】【fhq treap】A Simple Problem with Integers

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.

Source

POJ Monthly--2007.11.25, Yang Yi

【分析】

这个其实是个水题。我只是拿来fhq treap的...

Orzzzzzz范大爷

这种数据结构真是...能保证treap的性质同时也能保证合并后的树还能原封不动的切出来...

说它是treap，我觉得就一个fix像treap而已。

这个数据结构的精髓在合并和分裂(和打标记？)。有注释.

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <vector>
 #include <utility>
 #include <iomanip>
 #include <string>
 #include <cmath>
 #include <queue>
 #include <assert.h>
 #include <map>
 #include <ctime>
 #include <cstdlib>
 #define LOCAL
 const int MAXN =  + ;
 const int INF = 0x3f3f3f3f;
 const int SIZE = ;
 const int maxnode =  ;
 using namespace std;
 typedef long long ll;
 ll n, m;
 struct fhqTreap{
        struct Node{
               Node *l, *r;
               ll delta, sum;
               ll size, fix, val;
        }*root, *null, mem[];
        ll tot;

        ll BIG_RAND(){return (ll)rand()*RAND_MAX + (ll)rand();}
        void init(){
             tot = ;
             NEW(null, );
             null->size = ;
             root = null;
             insert(, n);//插入
        }
        void update(Node *t){
             if (t == null) return;
             t->size = t->l->size + t->r->size + ;
             t->sum = t->val;
             if (t->l != null) t->sum += t->l->sum;
             if (t->r != null) t->sum += t->r->sum;
        }
        //标记下传=-=
        void push_down(Node *t){
             if (t->delta){
                t->val += t->delta;

                if (t->l != null) {t->l->delta += t->delta, t->l->sum += t->l->size * t->delta;}
                if (t->r != null) {t->r->delta += t->delta, t->r->sum += t->r->size * t->delta;}

                t->delta = ;
             }
        }

        void NEW(Node *&t, ll val){
             t = &mem[tot++];
             t->size = ;
             t->fix = BIG_RAND();
             t->sum = t->val = val;
             t->delta = ;
             t->l = t->r = null;
        }
        //将t分裂为大小p和t->size - p的两个子树
        void split(Node *t, Node *&a, Node *&b, int p){
             if (t->size <= p) a = t, b = null;
             else if (p == ) a = null, b = t;
             else {
                  push_down(t);
                  if (t->l->size >= p){
                     b = t;
                     split(t->l, a , b->l, p);
                     update(b);
                  }else{
                     a = t;
                     split(t->r, a->r, b, p - t->l->size - );
                     update(a);
                  }
             }
        }
        //将a,b树合并为t
        void merge(Node *&t, Node *a, Node *b){
             if (a == null) t = b;
             else if (b == null) t = a;
             else {
                  if (a->fix > b->fix){//按fix值排序
                     push_down(a);
                     t = a;
                     merge(t->r, a->r, b);
                  }else{
                     push_down(b);
                     t = b;
                     merge(t->l, a, b->l);
                  }
                  update(t);
             }
        }
        void add(ll l, ll r, ll val){
             Node *a, *b, *c;
             split(root, a, b, l - );
             split(b, b, c, r - l + );
             b->delta += val;
             b->sum += val * b->size;
             merge(a, a, b);
             merge(root, a, c);
        }
        ll query(ll l, ll r){
           Node *a, *b, *c;
           split(root, a, b, l - );
           split(b, b, c, r - l + );
           ll ans = b->sum;
           merge(b, b, c);
           merge(root, a, b);
           return ans;
        }
        //把b挤出去
        void Delete(ll p){
             Node *a, *b, *c;
             split(root, a, b, p - );
             split(b, b, c, );
             merge(root, a, c);
        }
        Node *Insert(ll x){//不可思议的插入方式
             if (x == ) return null;
             if (x == ){
                ll tmp;
                scanf("%lld", &tmp);
                Node *a;
                NEW(a, tmp);
                return a;
             }
             Node *a, *b;
             int mid = x / ;
             a = Insert(mid);
             b = Insert(x - mid);
             merge(a, a, b);
             return a;
        } 

        //在pos处插入x个数字
        void insert(ll pos, ll x){
             Node *a, *b, *c, *t;
             //跟块状链表的有点像，分裂再合并
             split(root, a, b, pos);
             c = Insert(x);//插入一个值为x的树
             merge(a, a, c);
             merge(root, a, b);
        }
 }A;

 void work(){
      char str[];
      while (m--){
            scanf("%s", str);
            if (str[] == 'Q'){
               ll l, r;
               scanf("%lld%lld", &l, &r);
               printf("%lld\n", A.query(l, r));
            }else{
               ll l, r, k;
               scanf("%lld%lld%lld", &l, &r, &k);
               A.add(l, r, k);
            }
      }
 }

 int main(){

     while( scanf("%lld%lld", &n, &m) != EOF){
            A.init();
            work();
     }
     return ;
 }