【POJ】3468 A Simple Problem with Integers ——线段树 成段更新 懒惰标记

　A Simple Problem with Integers

Time Limit:5000MS   Memory Limit:131072K

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.

 

题解：本题也是线段树系列的模板题之一，要求的是成段更新+懒惰标记。PS：原题的说明有问题，上面说的“C a b c”中的c的范围其实在int32之外，需要使用long long，否则定是WA，真心坑爹，连WA多次，还是在discuss中看到的原因。

 

稍微讲解下代码中的一些细节: step<<1 与 step<<1|1，意思分别是step*2 和step*+1，具体为什么，可以回去复习一下位运算

 

AC代码如下：

 

 #include <cstdio>
 #include <cstring>

 typedef long long ll;
 const int LEN =  * ;

 struct line
 {
     int left;
     int right;
     ll value;
     ll lazy;  //懒惰标记
 }line[LEN];

 void buildt(int l, int r, int step)  //建树初始化
 {
     line[step].left = l;
     line[step].right = r;
     line[step].lazy = ;
     line[step].value = ;
     if (l == r)
         return;
     int mid = (l + r) / ;
     buildt(l, mid, step<<);
     buildt(mid+, r, step<<|);
 }

 void pushdown(int step)
 {
     if (line[step].left == line[step].right) //如果更新到最深处的子节点，返回
         return;
     if (line[step].lazy != ){  //如果有懒惰标记，向下传递懒惰标记且更新两个子节点的值
         line[step<<].lazy += line[step].lazy;
         line[step<<|].lazy += line[step].lazy;
         line[step<<].value += (line[step<<].right - line[step<<].left + ) * line[step].lazy;
         line[step<<|].value += (line[step<<|].right - line[step<<|].left + ) * line[step].lazy;
         line[step].lazy = ;
     }
 }

 void update(int l, int r, ll v, int step)
 {
     line[step].value += v * (r-l+);  //更新到当前节点，就在当前节点的value中加上增加的值
     pushdown(step);
     if (line[step].left == l && line[step].right == r){ //如果到达目标线段，做上懒惰标记，返回
         line[step].lazy = v;
         return;
     }
     int mid = (line[step].left + line[step].right) / ;
     if (r <= mid)
         update(l, r, v, step<<);
     else if (l > mid)
         update(l, r, v, step<<|);
     else{
         update(l, mid, v, step<<);
         update(mid+, r, v, step<<|);
     }
 }

 ll findans(int l, int r, int step)
 {
     if (l == line[step].left && r == line[step].right)  //如果找到目标线段，返回值
         return line[step].value;
     pushdown(step);
     int mid = (line[step].left + line[step].right) / ;
     if (r <= mid)
         return findans(l, r, step<<);
     else if (l > mid)
         return findans(l, r, step<<|);
     else
         return findans(l, mid, step<<) + findans(mid+, r, step<<|);
 }

 int main()
 {
     //freopen("in.txt", "r", stdin);
     int n, q;
     scanf("%d %d", &n, &q);
     buildt(, n, );
     for(int i = ; i <= n; i++){
         ll t;
         scanf("%I64d", &t);
         update(i, i, t, );
     }
     for(int i = ; i < q; i++){
         char query[];
         scanf("%s", query);
         if (query[] == 'C'){
             int a, b;
             ll c;
             scanf("%d %d %I64d", &a, &b, &c);
             update(a, b, c, );
         }
         else if (query[] == 'Q'){
             int a, b;
             scanf("%d %d", &a, &b);
             printf("%I64d\n", findans(a, b, ));
         }
     }
     return ;
 }