【Poj】-A Simple Problem with Integers（線段樹，區間更新）
阿新 • • 發佈：2019-01-23
A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 100877 | Accepted: 31450 | |
| 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 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.Source
區間更新的裸題，理解
#include<cstdio> #include<cstring> #include<stack> #include<cmath> #include<queue> #include<algorithm> using namespace std; #define CLR(a,b) memset(a,b,sizeof(a)) #define INF 0x3f3f3f3f #define LL long long #define MAX 100000 #define L o<<1 #define R o<<1|1 struct node { int l,r; LL sum,lazy; //延遲標記 int length; //區間長度 }Tree[MAX<<2]; void PushUp(int o) { Tree[o].sum=Tree[L].sum+Tree[R].sum; } void Build(int o,int l,int r) { Tree[o].l=l; Tree[o].r=r; Tree[o].lazy=0; Tree[o].length=r-l+1; if(l==r) { scanf("%lld",&Tree[o].sum); return ; } int mid=(Tree[o].l+Tree[o].r)>>1; Build(L,l,mid); Build(R,mid+1,r); PushUp(o); } void PushDown(int o) //向下傳遞 lazy ，o的lazy=0 { if(Tree[o].lazy) { Tree[L].sum+=Tree[L].length*Tree[o].lazy; Tree[R].sum+=Tree[R].length*Tree[o].lazy; Tree[L].lazy+=Tree[o].lazy; Tree[R].lazy+=Tree[o].lazy; Tree[o].lazy=0; } } void UpDate(int o,int l,int r,int lazy) { if(Tree[o].l==l&&Tree[o].r==r) { Tree[o].sum+=Tree[o].length*lazy; //更新當前節點的值，後向下傳遞 lazy Tree[o].lazy+=lazy; return ; } PushDown(o); int mid=(Tree[o].l+Tree[o].r)>>1; if(r<=mid) UpDate(L,l,r,lazy); else if(l>mid) UpDate(R,l,r,lazy); else { UpDate(L,l,mid,lazy); UpDate(R,mid+1,r,lazy); } PushUp(o); } LL Query(int o,int l,int r) { if(Tree[o].l==l&Tree[o].r==r) return Tree[o].sum; PushDown(o); int mid=(Tree[o].l+Tree[o].r)>>1; if(r<=mid) return Query(L,l,r); else if(l>mid) return Query(R,l,r); else return Query(L,l,mid)+Query(R,mid+1,r); } int main() { int n,m; scanf("%d%d",&n,&m); Build(1,1,n); char s[2]; int a,b,c; while(m--) { scanf("%s",s); if(s[0]=='Q') { scanf("%d%d",&a,&b); printf("%lld\n",Query(1,a,b)); } else { scanf("%d%d%d",&a,&b,&c); UpDate(1,a,b,c); } } return 0; }