【POJ3468】【zkw线段树】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】

很不错的数据结构，真的很不错....

跟树状数组有点像..

相比与普通线段树,zkw线段树具有标记永久化，简单好写的特点。

但zkw线段树占用的空间会相对大些，毕竟是2的阶乘...，然后应该不能可持久化吧(别跟我说用可持久化线段树维护可持久化数组....)

这样就不能动态分配内存了。。不过总体来说还是有一定的实用价值...

 /*
 宋代陆游
 《临安春雨初霁》

 世味年来薄似纱，谁令骑马客京华。
 小楼一夜听春雨，深巷明朝卖杏花。
 矮纸斜行闲作草，晴窗细乳戏分茶。
 素衣莫起风尘叹，犹及清明可到家。
 */
 #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>
 #include <stack>
 #define LOCAL
 const int INF = 0x7fffffff;
 const int MAXN =  + ;
 const int maxnode = ;
 using namespace std;
 struct Node{
     int l, r;
     long long sum, d;
 }tree[maxnode];
 int data[MAXN], M;//M为总结点个数        

 //初始化线段树
 void build(int l, int r){
     for (int i =  * M - ; i > ; i--){
         if (i >= M){
             tree[i].sum = data[i - M];
             tree[i].l = tree[i].r = (i - M);//叶子节点
         }
         else {
             tree[i].sum = tree[i<<].sum + tree[(i<<)+].sum;
             tree[i].l = tree[i<<].l;
             tree[i].r = tree[(i<<)+].r;
         }
     }
 }
 long long query(int l, int r){
     long long sum = ;//sum记录和
     int numl = , numr = ;
     l += M - ;
         r += M + ;
     while ((l ^ r) != ){
           if (~l&){//l取反，表示l是左节点
               sum += tree[l ^ ].sum;
               numl += (tree[l ^ ].r - tree[l ^ ].l + );
            }
            if (r & ){
               sum += tree[r ^ ].sum;
               numr += (tree[r ^ ].r - tree[r ^ ].l + );
            }
            l>>=;
            r>>=;
            //numl,numr使用来记录标记的
            sum += numl * tree[l].d;
            sum += numr * tree[r].d;
     }
     for (l >>= ; l > ; l >>= )  sum += (numl + numr) * tree[l].d;
     return sum;
 }

 void add(int l, int r, int val){
      int numl = , numr = ;
      l += M - ;
      r += M + ;
      while ((l ^ r) != ){
            if (~l&){//l取反
               tree[l ^ ].d += val;
               tree[l ^ ].sum += (tree[l ^ ].r - tree[l ^ ].l + ) * val;
               numl += (tree[l ^ ].r - tree[l ^ ].l + );
            }
            if (r & ){
               //更新另一边
               tree[r ^ ].d += val;
               tree[r ^ ].sum += (tree[r ^ ].r - tree[r ^ ].l + ) * val;
               numr += (tree[r ^ ].r - tree[r ^ ].l + );
            }
            l>>=;
            r>>=;
            tree[l].sum += numl * val;
            tree[r].sum += numr * val;
      }
      //不要忘了往上更新
      for (l >>= ; l > ; l >>= )  tree[l].sum += (numl+numr) * val;
 }
 int    n, m;

 void init(){
      scanf("%d%d", &n, &m);
      for (int i = ; i <= n; i++) scanf("%d", &data[i]);
      M = ;while (M < (n + )) M <<= ;//找到最大的段
      build(, M - );
 }
 void work(){
      while (m--){
             char str[];
             scanf("%s", str);
             if (str[] == 'Q'){
             int l, r;
             scanf("%d%d", &l, &r);
             printf("%lld\n", query(l, r));
         }
         else{
             int val, l, r;
             scanf("%d%d%d", &l, &r, &val);
             if (val != ) add(l, r, val);
         }
      }
 }

 int main(){

     init();
     work();
     return ;
 }