POJ 3468 A Simple Problem with Integers (线段树区域更新)
A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 62431 | Accepted: 19141 | |
| 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
POJ Monthly--2007.11.25, Yang Yi
线段树的基础区域更新
#include<cstdio> #include<iostream> #include<cstring> #include<stack> #include<queue> #include<cmath> #include<stdlib.h> #include<algorithm> #define LL __int64 using namespace std; const int MAXN=100000+5; const int INF=0x3f3f3f3f; struct node { int l,r; LL num,flag; int mid() { return (l+r)>>1; } }a[MAXN*4]; int b[MAXN],n,m; void pushdown(int len,int step) { if(a[step].flag) { a[step*2].flag+=a[step].flag; a[step*2+1].flag+=a[step].flag; a[step*2].num+=(len-len/2)*a[step].flag; a[step*2+1].num+=(len/2)*a[step].flag; a[step].flag=0; } } void build(int l,int r,int step) { a[step].l=l; a[step].r=r; a[step].flag=0; if(l==r) { a[step].num=b[l]; return ; } int mid=a[step].mid(); build(l,mid,step*2); build(mid+1,r,step*2+1); a[step].num=a[step*2].num+a[step*2+1].num; } void update(int x,int y,LL val,int step) { if(x<=a[step].l && a[step].r<=y) { a[step].flag+=val; a[step].num+=(a[step].r-a[step].l+1)*val; return ; } pushdown(a[step].r-a[step].l+1,step); int mid=a[step].mid(); if(x>mid) update(x,y,val,step*2+1); else if(y<=mid) update(x,y,val,step*2); else { update(x,mid,val,step*2); update(mid+1,y,val,step*2+1); } a[step].num=a[step*2].num+a[step*2+1].num; } LL query(int x,int y,int step) { if(x<=a[step].l && a[step].r<=y) return a[step].num; pushdown(a[step].r-a[step].l+1,step); int mid=a[step].mid(); if(x>mid) return query(x,y,step*2+1); else if(y<=mid) return query(x,y,step*2); else return query(x,y,step*2)+query(x,y,step*2+1); } int main() { while(scanf("%d %d",&n,&m)!=EOF) { for(int i=1;i<=n;i++) scanf("%d",&b[i]); build(1,n,1); while(m--) { char str[10]; int A,B; scanf("%s %d %d",str,&A,&B); if(str[0]=='Q') printf("%I64d\n",query(A,B,1)); if(str[0]=='C') { int cnt; scanf("%d",&cnt); update(A,B,cnt,1); } } } return 0; }