線段樹入門 理解

#include <iostream>#include <cmath>#include <cstring>#include <algorithm>using namespace std;int n,m;int a[100001]; // 葉子結點數N__int64 ans;struct{ int l,r; __int64 sum,add; //__int64 }tree[400000]; //數中所有結點數量大概為 2^2*N 層數logN  h=logN+1 或+2  每次插入更新不超過 O(logN)void build(int l,int r,int i){ tree[i].l = l; tree[i].r = r; tree[i].add = 0; if(l == r) {  tree[i].sum = a[l]; //葉子結點的值=輸入數組的值  return ; } int mid = (l+r)/2; build(l,mid,2*i); build(mid+1,r,2*i+1); tree[i].sum = tree[2*i].sum + tree[2*i+1].sum;}void updata(int l,int r,int add,int i){ if(tree[i].l > r || tree[i].r < l)  return ; if(tree[i].l >= l && tree[i].r <= r) {  tree[i].sum += (tree[i].r-tree[i].l+1)*add;  tree[i].add += add;  return ; //lazy 只加到分解唯一的子區間 } if(tree[i].add) //lazy 邊更新邊維護 為了確保 孩子的 add和 sum正確 必須更新 且更新完後 當前add=0 （更新到最底）的意思  {  tree[2*i].sum += (tree[2*i].r-tree[2*i].l+1)*tree[i].add;  tree[2*i].add += tree[i].add;  tree[2*i+1].sum += (tree[2*i+1].r-tree[2*i+1].l+1)*tree[i].add;  tree[2*i+1].add += tree[i].add;  tree[i].add = 0; } updata(l,r,add,2*i); updata(l,r,add,2*i+1); tree[i].sum = tree[2*i].sum + tree[2*i+1].sum;}void query(int l,int r,int i){ if(tree[i].l > r || tree[i].r < l)  return ; if(tree[i].l >= l && tree[i].r <= r) {  ans += tree[i].sum;  return ; } if(tree[i].add) //lazy 邊查詢邊維護  {  tree[2*i].sum += (tree[2*i].r-tree[2*i].l+1)*tree[i].add;  tree[2*i].add += tree[i].add;  tree[2*i+1].sum += (tree[2*i+1].r-tree[2*i+1].l+1)*tree[i].add;  tree[2*i+1].add += tree[i].add;  tree[i].add = 0; } query(l,r,2*i); query(l,r,2*i+1); tree[i].sum = tree[2*i].sum + tree[2*i+1].sum;}int main(){ //freopen("in.txt","r",stdin); int i,x,y,z; char c; while(scanf("%d%d",&n,&m)!=EOF) {  for(i=1;i<=n;i++)   scanf("%d",&a[i]);  build(1,n,1);  scanf("%*c");  for(i=1;i<=m;i++)  {   scanf("%c",&c);   if(c == 'Q')   {    scanf("%d%d%*c",&x,&y);    ans = 0;    query(x,y,1);    printf("%I64d\n",ans);   }   else   {    scanf("%d%d%d%*c",&x,&y,&z);    updata(x,y,z,1);   }  } } return 0;}

扒了一線段樹代碼 注釋作分析