【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：

 

 

  1 #include <cstdio>

  2 #include <cstring>

  3 

  4 typedef long long ll;

  5 const int LEN = 100000 * 4;

  6 

  7 struct line

  8 {

  9     int left;

 10     int right;

 11     ll value;

 12     ll lazy;  //懒惰标记

 13 }line[LEN];

 14 

 15 void buildt(int l, int r, int step)  //建树初始化

 16 {

 17     line[step].left = l;

 18     line[step].right = r;

 19     line[step].lazy = 0;

 20     line[step].value = 0;

 21     if (l == r)

 22         return;

 23     int mid = (l + r) / 2;

 24     buildt(l, mid, step<<1);

 25     buildt(mid+1, r, step<<1|1);

 26 }

 27 

 28 void pushdown(int step)

 29 {

 30     if (line[step].left == line[step].right) //如果更新到最深处的子节点，返回

 31         return;

 32     if (line[step].lazy != 0){  //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

 33         line[step<<1].lazy += line[step].lazy;

 34         line[step<<1|1].lazy += line[step].lazy;

 35         line[step<<1].value += (line[step<<1].right - line[step<<1].left + 1) * line[step].lazy;

 36         line[step<<1|1].value += (line[step<<1|1].right - line[step<<1|1].left + 1) * line[step].lazy;

 37         line[step].lazy = 0;

 38     }

 39 }

 40 

 41 void update(int l, int r, ll v, int step)

 42 {

 43     line[step].value += v * (r-l+1);  //更新到当前节点，就在当前节点的value中加上增加的值

 44     pushdown(step);

 45     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

 46         line[step].lazy = v;

 47         return;

 48     }

 49     int mid = (line[step].left + line[step].right) / 2;

 50     if (r <= mid)

 51         update(l, r, v, step<<1);

 52     else if (l > mid)

 53         update(l, r, v, step<<1|1);

 54     else{

 55         update(l, mid, v, step<<1);

 56         update(mid+1, r, v, step<<1|1);

 57     }

 58 }

 59 

 60 ll findans(int l, int r, int step)

 61 {

 62     if (l == line[step].left && r == line[step].right)  //如果找到目标线段，返回值

 63         return line[step].value;

 64     pushdown(step);

 65     int mid = (line[step].left + line[step].right) / 2;

 66     if (r <= mid)

 67         return findans(l, r, step<<1);

 68     else if (l > mid)

 69         return findans(l, r, step<<1|1);

 70     else

 71         return findans(l, mid, step<<1) + findans(mid+1, r, step<<1|1);

 72 }

 73 

 74 int main()

 75 {

 76     //freopen("in.txt", "r", stdin);

 77     int n, q;

 78     scanf("%d %d", &n, &q);

 79     buildt(1, n, 1);

 80     for(int i = 1; i <= n; i++){

 81         ll t;

 82         scanf("%I64d", &t);

 83         update(i, i, t, 1);

 84     }

 85     for(int i = 0; i < q; i++){

 86         char query[2];

 87         scanf("%s", query);

 88         if (query[0] == 'C'){

 89             int a, b;

 90             ll c;

 91             scanf("%d %d %I64d", &a, &b, &c);

 92             update(a, b, c, 1);

 93         }

 94         else if (query[0] == 'Q'){

 95             int a, b;

 96             scanf("%d %d", &a, &b);

 97             printf("%I64d\n", findans(a, b, 1));

 98         }

 99     }

100     return 0;

101 }