POJ-3468 A Simple Problem with Integers ( 线段树 )
题目链接: http://poj.org/problem?id=3468
Description
You have N integers, A1, A2, ... , AN. 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, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.
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
询问区间和并对区间进行操作。
对区间操作是不能点更新，应该区间更新，不然会超时 例如区间(1,10)，我要对(3,8)进行操作，点更新需要进行(1,5),(1,3),(3,3),(4,5),(4,4),(5,5),(6,10),(6,8),(6,7),(6,6),(7,7),(8,8)共12步，而区间操作只需要(1,5),(1,3),(3,3),(4,5),(6,10),(6,8)共6步，而这只是一种简单的情况。
进行区间操作需要在节点加入一个参数add，用来记录该区间每个元素的增值以避免点更新。
1 #include<iostream> 2 #include<cstring> 3 using namespace std; 4 5 class node{ 6 public: 7 int l, r; 8 long long v, add; //v为该区间的和，并用add记录加在该区间上的值 9 node *left, *right; 10 }; 11 long long num[100005]; 12 13 node *Creat( int z, int y ){ 14 node *root = new node; 15 root->l = z; 16 root->r = y; 17 root->add = 0; 18 root->left = NULL; 19 root->right = NULL; 20 if( z == y ){ 21 root->v = num[z]; 22 return root; 23 } 24 25 int mid = ( z + y ) >> 1; 26 root->left = Creat( z, mid ); 27 root->right = Creat( mid + 1, y ); 28 29 root->v = root->left->v + root->right->v; 30 31 return root; 32 } 33 34 void Insert( int z, int y, long long n, node *p ){ 35 if( z == p->l && y == p->r ){ 36 p->add += n; 37 return; 38 } 39 40 p->v += ( y - z + 1 ) * n; 41 42 int mid = ( p->l + p->r ) >> 1; 43 44 if( y <= mid ) Insert( z, y, n, p->left ); 45 else if( z > mid ) Insert( z, y, n, p->right ); 46 else{ 47 Insert( z, mid, n, p->left ); 48 Insert( mid +1, y, n, p->right ); 49 } 50 } 51 52 long long Ask( int z, int y,node *p ){ 53 long long ans = 0; 54 55 if( z == p->l && y == p->r ) 56 return p->v + p->add * ( y - z +1 ); 57 58 ans += p->add * ( y - z + 1 ); 59 60 int mid = ( p->l + p->r ) >> 1; 61 62 if( y <= mid ) ans += Ask( z, y, p->left ); 63 else if( z > mid ) ans += Ask( z, y, p->right ); 64 else 65 ans += Ask( z, mid, p->left ) + Ask( mid + 1, y, p->right ); 66 67 return ans; 68 } 69 70 int main(){ 71 ios::sync_with_stdio( false ); 72 73 int N, Q; 74 cin >> N >> Q; 75 76 for( int i = 1; i <= N; i++ ) 77 cin >> num[i]; 78 79 node *root = Creat( 1, N ); 80 81 int a, b; 82 long long c; 83 char k; 84 for( int i = 1; i <= Q; i++ ){ 85 cin >> k; 86 87 if( k == 'C' ){ 88 cin >> a >> b >> c; 89 Insert( a, b, c, root); 90 } 91 92 else{ 93 cin >> a >> b; 94 cout << Ask( a, b, root ) << endl; 95 } 96 } 97 98 return 0; 99 }
