线段树（1）
线段树如图，图要记清楚，就是一棵饱和二叉树，最后一排没有画的的也存在，只是没有值，所以左节点为2u,右节点为2u+1，所以对于区间【1，n】,结构体数组应为4n.
对于线段树基础，有几个主要的模块。
1，建立线段树（初始Build（））
1 void build(int u,int left,int right) 2 { 3 node[u].l=left; 4 node[u].r=right; 5 node[u].add=0; 6 if(node[u].l==node[u].r) 7 { 8 node[u].sum=a[left]; 9 return ;//叶子节点 10 } 11 int mid=(left+right)>>1; 12 build(u*2,left,mid); 13 build(u*2+1,mid+1,right); 14 Pushup(u);//只赋值了叶子节点，向上更新 15 }
2，更新操作（加一个值，乘一个字Updata()）
1 void updata(int u,int left,int right,int ad) 2 { 3 if(node[u].l==left&&node[u].r==right) 4 { 5 node[u].add+=ad;//有些还没有pushdouwn的.add要加上去一起更新 6 node[u].sum+=(right-left+1)*ad; 7 return; 8 } 9 node[u].sum+=(right-left+1)*ad;//u节点区间包含[left,right],也要加上 10 if(node[u].add)pushdown(u);//延时向下更新 11 int mid=(node[u].l+node[u].r)>>1; 12 if(right<=mid)updata(u*2,left,right,ad); 13 else if(left>mid)updata(u*2+1,left,right,ad);//若左右边都有，分成两段计算 14 else 15 { 16 updata(u*2,left,mid,ad); 17 updata(u*2+1,mid+1,right,ad);//同时占有左右子树，把他分成两部分计算 18 } 19 // pushup(u);因为有node[u].sum+=(right-left+1)*ad; 20 }
3，查询（一段区间的和Query（））
1 long long Query(int u,int left,int right) 2 { 3 if(node[u].l==left&&node[u].r==right) 4 return node[u].sum; 5 if(node[u].add)pushdown(u);//add不为0，说明还未更新，查询就要顺便向下更新(便于延时更新) 6 int mid=(node[u].l+node[u].r)>>1; 7 if(right<=mid)return Query(u*2,left,right); 8 else if(left>mid)return Query(u*2+1,left,right); 9 else 10 return (Query(u*2,left,mid)+Query(u*2+1,mid+1,right)); 11 12 //pushup(u);//同理 13 }
4，向上回溯（Pushup（））
1 void Pushup(int u) 2 { 3 node[u].sum=node[u*2].sum+node[u*2+1].sum; 4 return ; 5 }
5，向下延时更新（Pushdown（））----为什么是延时，看下面
1 void pushdown(int u) 2 { 3 node[u*2].add+=node[u].add; 4 node[u*2].sum+=(node[u*2].r-node[u*2].l+1)*node[u].add; 5 node[u*2+1].add+=node[u].add; 6 node[u*2+1].sum+=(node[u*2+1].r-node[u*2+1].l+1)*node[u].add; 7 node[u].add=0;//加完就清0，说明已经更新过了 8 }
poj 3468 A Simple Problem with Integers（线段树）
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
代码
1 #include<iostream> 2 #include<cstdio> 3 #define N 100005 4 using namespace std; 5 int n,m; 6 int a[N]; 7 struct Node 8 { 9 int l,r;//节点所包含的左右区间端点 10 long long sum; 11 long long add;//add是指每个所增加的（即为C） 12 }node[4*N]; 13 void Pushup(int u) 14 { 15 node[u].sum=node[u*2].sum+node[u*2+1].sum; 16 return ; 17 } 18 void pushdown(int u) 19 { 20 node[u*2].add+=node[u].add; 21 node[u*2].sum+=(node[u*2].r-node[u*2].l+1)*node[u].add; 22 node[u*2+1].add+=node[u].add; 23 node[u*2+1].sum+=(node[u*2+1].r-node[u*2+1].l+1)*node[u].add; 24 node[u].add=0;//加完就清0，说明已经更新过了 25 } 26 void build(int u,int left,int right) 27 { 28 node[u].l=left; 29 node[u].r=right; 30 node[u].add=0; 31 if(node[u].l==node[u].r) 32 { 33 node[u].sum=a[left]; 34 return ;//叶子节点 35 } 36 int mid=(left+right)>>1; 37 build(u*2,left,mid); 38 build(u*2+1,mid+1,right); 39 Pushup(u);//只赋值了叶子节点，向上更新 40 } 41 void updata(int u,int left,int right,int ad) 42 { 43 if(node[u].l==left&&node[u].r==right) 44 { 45 node[u].add+=ad;//有些还没有pushdouwn的.add要加上去一起更新 46 node[u].sum+=(right-left+1)*ad; 47 return; 48 } 49 node[u].sum+=(right-left+1)*ad;//u节点区间包含[left,right],也要加上 50 if(node[u].add)pushdown(u);//延时向下更新 51 int mid=(node[u].l+node[u].r)>>1; 52 if(right<=mid)updata(u*2,left,right,ad); 53 else if(left>mid)updata(u*2+1,left,right,ad);//若左右边都有，分成两段计算 54 else 55 { 56 updata(u*2,left,mid,ad); 57 updata(u*2+1,mid+1,right,ad);//同时占有左右子树，把他分成两部分计算 58 } 59 // pushup(u);因为有node[u].sum+=(right-left+1)*ad; 60 } 61 long long Query(int u,int left,int right) 62 { 63 if(node[u].l==left&&node[u].r==right) 64 return node[u].sum; 65 if(node[u].add)pushdown(u);//add不为0，说明还未更新，查询就要顺便向下更新 66 int mid=(node[u].l+node[u].r)>>1; 67 if(right<=mid)return Query(u*2,left,right); 68 else if(left>mid)return Query(u*2+1,left,right); 69 else 70 return (Query(u*2,left,mid)+Query(u*2+1,mid+1,right)); 71 72 //pushup(u);//同理 73 } 74 int main() 75 { 76 freopen("POJ3468.in","r",stdin); 77 freopen("POJ3468.out","w",stdout); 78 while((scanf("%d%d",&n,&m))==2) 79 { 80 for(int i=1;i<=n;i++) 81 scanf("%d",&a[i]); 82 getchar(); 83 build(1,1,n); 84 for(int i=1;i<=m;i++) 85 { 86 int x,y,z; 87 char c; 88 scanf("%c ",&c); 89 if(c=='Q') 90 { 91 scanf("%d%d",&x,&y); 92 long long w=Query(1,x,y); 93 cout<<w<<endl; 94 getchar(); 95 } 96 if(c=='C') 97 { 98 scanf("%d%d%d",&x,&y,&z); 99 updata(1,x,y,z); 100 getchar(); 101 } 102 103 } 104 } 105 106 return 0; 107 }