POJ 3468 A Simple Problem with Integers //線段樹的成段更新

標籤：線段樹

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 59046		Accepted: 17974
Case Time Limit: 2000MS

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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

Hint

The sums may exceed the range of 32-bit integers.

Source

POJ Monthly--2007.11.25, Yang Yi

/*區間更新的lazy操作。*/#include <stdio.h>struct node{int l, r;__int64 sum;__int64 lazy;//當成段更新時，往往不用更新到單個的點。lazy操作大大節省了時間。}tree[300005];int h[100005];__int64 sum;//int超限void build(int l, int r, int n){int mid;tree[n].l = l;tree[n].r = r;tree[n].lazy = 0;//賦初值if(l==r){tree[n].sum = h[l];return ;}mid = (l+r)/2;build(l, mid, 2*n);build(mid+1, r, 2*n+1);tree[n].sum = tree[2*n].sum+tree[2*n+1].sum;}void add(int l, int r, __int64 k, int n){int mid;if(tree[n].l==l && tree[n].r==r)//當需要更新的段 與 結點對應的段吻合時，直接把此結點的lazy值更新即可，不需要再向下更新。{tree[n].lazy += k;return;}tree[n].sum += k*(r-l+1);//當此區間包含需要更新的區間，但不吻合時，需要向下繼續尋找，此時需要更新這個父節點的sum值。mid = (tree[n].l + tree[n].r)/2;if(r <= mid)add(l, r, k, 2*n);else if(l >=mid+1)add(l, r, k, 2*n+1);else{add(l, mid, k, 2*n);add(mid+1, r, k, 2*n+1);}}void qu(int l, int r, int n){int mid;if(tree[n].l==l && tree[n].r==r){sum += tree[n].sum + (r-l+1)*tree[n].lazy;//當尋找的段與 此結點的段吻合時，sum 值等於這個結點的sum加上lazy乘區間長度的值。return ;}if(tree[n].lazy!=0 && tree[n].l!=tree[n].r)//當尋找區間為此結點對應區間的子集時，需要將此結點對應的lazy值下放到其子節點，並把此結點的lazy值置為0。{add(tree[2*n].l, tree[2*n].r, tree[n].lazy, n);add(tree[2*n+1].l, tree[2*n+1].r, tree[n].lazy, n);tree[n].lazy = 0;}mid = (tree[n].l + tree[n].r)/2;if(l >= mid+1)qu(l, r, 2*n+1);else if(r <= mid)qu(l, r, 2*n);else{qu(l, mid, 2*n);qu(mid+1, r, 2*n+1);}}int main(){int n, q;int i;int a, b, c;char ch[10];scanf("%d%d", &n, &q);for(i=1; i<=n; i++)scanf("%d", &h[i]);build(1, n, 1);while(q--){scanf("%s", ch);if(ch[0]=='Q'){scanf("%d%d", &a, &b);sum = 0;qu(a, b, 1);printf("%I64d\n", sum);}else{scanf("%d%d%d", &a, &b, &c);add(a, b, c, 1);}}return 0;}